What Is Physics?

One job of physics is to identify the different types of energy in the world,

especially those that are of common importance. One general type of energy is

potential energy U. Technically, potential energy is energy that can be associated

with the configuration (arrangement) of a system of objects that exert forces on

one another.

CHAPTER 8

Potential Energy and

Conservation of Energy

8-1 POTENTIAL ENERGY

After reading this module, you should be able to . . .

8.01 Distinguish a conservative force from a nonconservative

force.

8.02 For a particle moving between two points, identify that

the work done by a conservative force does not depend on

which path the particle takes.

8.03 Calculate the gravitational potential energy of a particle

(or, more properly, a particle–Earth system).

8.04 Calculate the elastic potential energy of a block–spring

system.

●A force is a conservative force if the net work it does on

a particle moving around any closed path, from an initial

point and then back to that point, is zero. Equivalently, a force

is conservative if the net work it does on a particle moving

between two points does not depend on the path taken by

the particle. The gravitational force and the spring force are

conservative forces; the kinetic frictional force is a noncon-

servative force.

●Potential energy is energy that is associated with the con-

figuration of a system in which a conservative force acts.

When the conservative force does work Won a particle

within the system, the change Uin the potential energy of

the system is

UW.

If the particle moves from point xito point xf, the change in

the potential energy of the system is

U

xf

F(x)dx.

●The potential energy associated with a system consisting of

Earth and a nearby particle is gravitational potential energy. If

the particle moves from height yito height yf, the change in the

gravitational potential energy of the particle–Earth system is

Umg(yfyi)mg y.

●If the reference point of the particle is set as yi0and the

corresponding gravitational potential energy of the system is

set as Ui0, then the gravitational potential energy Uwhen

the particle is at any height yis

U(y)mgy.

●Elastic potential energy is the energy associated with the

state of compression or extension of an elastic object. For a

spring that exerts a spring force Fkx when its free end

has displacement x, the elastic potential energy is

●The reference configuration has the spring at its relaxed

length, at which x0and U0.

U(x)1

2kx2.

Key Ideas

Learning Objectives

177177

This is a pretty formal definition of something that is actually familiar to you.

An example might help better than the definition: A bungee-cord jumper plunges

from a staging platform (Fig. 8-1). The system of objects consists of Earth and the

jumper.The force between the objects is the gravitational force.The configuration

of the system changes (the separation between the jumper and Earth decreases—

that is, of course, the thrill of the jump). We can account for the jumper’s motion

and increase in kinetic energy by defining a gravitational potential energy U. This

is the energy associated with the state of separation between two objects that at-

tract each other by the gravitational force, here the jumper and Earth.

When the jumper begins to stretch the bungee cord near the end of the

plunge, the system of objects consists of the cord and the jumper. The force

between the objects is an elastic (spring-like) force. The configuration of the sys-

tem changes (the cord stretches). We can account for the jumper’s decrease in

kinetic energy and the cord’s increase in length by defining an elastic potential

energy U.This is the energy associated with the state of compression or extension

of an elastic object, here the bungee cord.

Physics determines how the potential energy of a system can be calculated so

that energy might be stored or put to use. For example, before any particular

bungee-cord jumper takes the plunge, someone (probably a mechanical engi-

neer) must determine the correct cord to be used by calculating the gravitational

and elastic potential energies that can be expected.Then the jump is only thrilling

and not fatal.

Work and Potential Energy

In Chapter 7 we discussed the relation between work and a change in kinetic energy.

Here we discuss the relation between work and a change in potential energy.

Let us throw a tomato upward (Fig. 8-2).We already know that as the tomato

rises, the work Wgdone on the tomato by the gravitational force is negative

because the force transfers energy from the kinetic energy of the tomato.We can

now finish the story by saying that this energy is transferred by the gravitational

force to the gravitational potential energy of the tomato–Earth system.

The tomato slows, stops, and then begins to fall back down because of the

gravitational force. During the fall, the transfer is reversed:The work Wgdone on

the tomato by the gravitational force is now positive—that force transfers energy

from the gravitational potential energy of the tomato–Earth system to the

kinetic energy of the tomato.

For either rise or fall, the change Uin gravitational potential energy is

defined as being equal to the negative of the work done on the tomato by the

gravitational force. Using the general symbol Wfor work, we write this as

UW. (8-1)

178 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Figure 8-1 The kinetic energy of a bungee-

cord jumper increases during the free fall,

and then the cord begins to stretch, slow-

ing the jumper.

Rough Guides/Greg Roden/Getty Images, Inc.

Figure 8-2 A tomato is thrown upward. As it rises, the

gravitational force does negative work on it, decreasing

its kinetic energy. As the tomato descends, the

gravitational force does positive work on it, increasing

its kinetic energy.

Negative

ork done

by the

gravitational

force

Positive

work done

by the

gravitational

force

This equation also applies to a block–spring system, as in Fig. 8-3. If we

abruptly shove the block to send it moving rightward, the spring force acts leftward

and thus does negative work on the block, transferring energy from the kinetic

energy of the block to the elastic potential energy of the spring–block system.The

block slows and eventually stops, and then begins to move leftward because the

spring force is still leftward. The transfer of energy is then reversed—it is from

potential energy of the spring–block system to kinetic energy of the block.

Conservative and Nonconservative Forces

Let us list the key elements of the two situations we just discussed:

1. The system consists of two or more objects.

2. Aforce acts between a particle-like object (tomato or block) in the system and

the rest of the system.

3. When the system configuration changes, the force does work (call it W1) on

the particle-like object, transferring energy between the kinetic energy Kof

the object and some other type of energy of the system.

4. When the configuration change is reversed, the force reverses the energy

transfer,doing work W2in the process.

In a situation in which W1W2is always true, the other type of energy is

a potential energy and the force is said to be a conservative force. As you might

suspect, the gravitational force and the spring force are both conservative (since

otherwise we could not have spoken of gravitational potential energy and elastic

potential energy, as we did previously).

A force that is not conservative is called a nonconservative force. The kinetic

frictional force and drag force are nonconservative. For an example, let us send

a block sliding across a floor that is not frictionless. During the sliding, a kinetic

frictional force from the floor slows the block by transferring energy from its

kinetic energy to a type of energy called thermal energy (which has to do with the

random motions of atoms and molecules). We know from experiment that this

energy transfer cannot be reversed (thermal energy cannot be transferred back

to kinetic energy of the block by the kinetic frictional force). Thus, although we

have a system (made up of the block and the floor), a force that acts between

parts of the system, and a transfer of energy by the force, the force is not conser-

vative.Therefore, thermal energy is not a potential energy.

When only conservative forces act on a particle-like object, we can greatly

simplify otherwise difficult problems involving motion of the object. Let’s next

develop a test for identifying conservative forces, which will provide one means

for simplifying such problems.

Path Independence of Conservative Forces

The primary test for determining whether a force is conservative or nonconserva-

tive is this: Let the force act on a particle that moves along any closed path, begin-

ning at some initial position and eventually returning to that position (so that the

particle makes a round trip beginning and ending at the initial position). The

force is conservative only if the total energy it transfers to and from the particle

during the round trip along this and any other closed path is zero. In other words:

179

8-1 POTENTIAL ENERGY

The net work done by a conservative force on a particle moving around any

closed path is zero.

Figure 8-3 A block, attached to a spring and

initially at rest at x0, is set in motion

toward the right. (a) As the block moves

rightward (as indicated by the arrow), the

spring force does negative work on it.

(b) Then, as the block moves back toward

x0, the spring force does positive work

on it.

(a)

(b)

We know from experiment that the gravitational force passes this closed-

path test. An example is the tossed tomato of Fig. 8-2. The tomato leaves the

launch point with speed v0and kinetic energy .The gravitational force acting

2mv0

Checkpoint 1

The figure shows three paths connecting points a

and b.A single force does the indicated work on

a particle moving along each path in the indicated

direction. On the basis of this information, is force

conservative?

on the tomato slows it, stops it, and then causes it to fall back down. When the

tomato returns to the launch point, it again has speed v0and kinetic energy

Thus, the gravitational force transfers as much energy from the tomato dur-

ing the ascent as it transfers to the tomato during the descent back to the launch

point. The net work done on the tomato by the gravitational force during the

round trip is zero.

An important result of the closed-path test is that:

2mv0

180 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

(a)

(b)

The force is

conservative. Any

choice of path

between the points

gives the same

amount of work.

And a round trip

gives a total work

of zero.

Figure 8-4 (a) As a conservative force acts

on it, a particle can move from point ato

point balong either path 1 or path 2.

(b) The particle moves in a round trip,

from point ato point balong path 1 and

then back to point aalong path 2.

–

60 J

The work done by a conservative force on a particle moving between two points

does not depend on the path taken by the particle.

For example, suppose that a particle moves from point ato point bin Fig. 8-4a

along either path 1 or path 2. If only a conservative force acts on the particle,then

the work done on the particle is the same along the two paths. In symbols, we can

write this result as

Wab,1 Wab,2, (8-2)

where the subscript ab indicates the initial and final points, respectively, and the

subscripts 1 and 2 indicate the path.

This result is powerful because it allows us to simplify difficult problems

when only a conservative force is involved. Suppose you need to calculate the

work done by a conservative force along a given path between two points, and

the calculation is difficult or even impossible without additional information.

You can find the work by substituting some other path between those two points

for which the calculation is easier and possible.

Proof of Equation 8-2

Figure 8-4bshows an arbitrary round trip for a particle that is acted upon by a single

force. The particle moves from an initial point ato point balong path 1 and then

back to point aalong path 2. The force does work on the particle as the particle

moves along each path. Without worrying about where positive work is done and

where negative work is done, let us just represent the work done from ato balong

path 1 as Wab,1 and the work done from bback to aalong path 2 as Wba,2. If the force

is conservative,then the net work done during the round trip must be zero:

Wab,1 Wba,2 0,

and thus

Wab,1 Wba,2. (8-3)

In words, the work done along the outward path must be the negative of the work

done along the path back.

Let us now consider the work Wab,2 done on the particle by the force when

the particle moves from ato balong path 2, as indicated in Fig. 8-4a. If the force is

conservative, that work is the negative of Wba,2:

Wab,2 Wba,2. (8-4)

Substituting Wab,2 for Wba,2 in Eq. 8-3, we obtain

Wab,1 Wab,2,

which is what we set out to prove.

Determining Potential Energy Values

Here we find equations that give the value of the two types of potential energy

discussed in this chapter: gravitational potential energy and elastic potential

energy. However, first we must find a general relation between a conservative

force and the associated potential energy.

Consider a particle-like object that is part of a system in which a conservative

force acts. When that force does work Won the object, the change Uin

the potential energy associated with the system is the negative of the work done.

We wrote this fact as Eq. 8-1 (UW). For the most general case, in which the

force may vary with position, we may write the work Was in Eq. 7-32:

(8-5)

This equation gives the work done by the force when the object moves from

point xito point xf, changing the configuration of the system. (Because the

force is conservative, the work is the same for all paths between those two

points.)

Wxf

F(x)dx.

181

8-1 POTENTIAL ENERGY

Sample Problem 8.01 Equivalent paths for calculating work, slippery cheese

The main lesson of this sample problem is this: It is perfectly

all right to choose an easy path instead of a hard path.

Figure 8-5ashows a 2.0 kg block of slippery cheese that

slides along a frictionless track from point ato point b.The

cheese travels through a total distance of 2.0 m along the

track, and a net vertical distance of 0.80 m. How much work is

done on the cheese by the gravitational force during the slide?

KEY IDEAS

(1) We cannot calculate the work by using Eq. 7-12 (Wg

mgd cos f). The reason is that the angle fbetween the

directions of the gravitational force and the displacement

varies along the track in an unknown way. (Even if we did

know the shape of the track and could calculate falong it,

the calculation could be very difficult.) (2) Because is a

conservative force, we can find the work by choosing some

other path between aand b—one that makes the calcula-

tion easy.

Calculations: Let us choose the dashed path in Fig. 8-5b;it

consists of two straight segments. Along the horizontal seg-

ment, the angle fis a constant 90. Even though we do not

know the displacement along that horizontal segment, Eq. 7-12

tells us that the work Whdone there is

Whmgd cos 900.

Along the vertical segment, the displacement dis 0.80 m

and, with and both downward, the angle fis a constantd

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(

)(

)

The gravitational force is conservative.

Any choice of path between the points

gives the same amount of work.

Figure 8-5 (a) A block of cheese slides along a frictionless track

from point ato point b.(b) Finding the work done on the cheese by

the gravitational force is easier along the dashed path than along

the actual path taken by the cheese; the result is the same for

both paths.

vertical part of the dashed path,

Wvmgd cos 0

(2.0 kg)(9.8 m/s2)(0.80 m)(1) 15.7 J.

The total work done on the cheese by as the cheese

moves from point ato point balong the dashed path is then

WWhWv015.7 J 16 J. (Answer)

This is also the work done as the cheese slides along the

track from ato b.

0. Thus, Eq. 7-12 gives us, for the work Wvdone along the

Substituting Eq. 8-5 into Eq. 8-1, we find that the change in potential energy

due to the change in configuration is, in general notation,

(8-6)

Gravitational Potential Energy

We first consider a particle with mass mmoving vertically along a yaxis (the

positive direction is upward). As the particle moves from point yito point yf,

the gravitational force does work on it.To find the corresponding change in

the gravitational potential energy of the particle–Earth system, we use Eq. 8-6

with two changes: (1) We integrate along the yaxis instead of the xaxis, because

the gravitational force acts vertically. (2) We substitute mg for the force symbol F,

because has the magnitude mg and is directed down the yaxis.We then have

which yields

Umg(yfyi)mg y. (8-7)

Only changes Uin gravitational potential energy (or any other type of

potential energy) are physically meaningful. However, to simplify a calculation or

a discussion, we sometimes would like to say that a certain gravitational potential

value Uis associated with a certain particle–Earth system when the particle is at

a certain height y.To do so, we rewrite Eq. 8-7 as

UUimg(yyi). (8-8)

Then we take Uito be the gravitational potential energy of the system when it is

in a reference configuration in which the particle is at a reference point yi.

Usually we take Ui0 and yi0. Doing this changes Eq. 8-8 to

U(y)mgy (gravitational potential energy). (8-9)

This equation tells us:

U

yf

(mg)dy mg yf

dy mg





U

xf

F(x)dx.

182 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Elastic Potential Energy

We next consider the block–spring system shown in Fig. 8-3, with the block

moving on the end of a spring of spring constant k. As the block moves from

point xito point xf, the spring force Fxkx does work on the block.To find the

corresponding change in the elastic potential energy of the block–spring system,

we substitute kx for F(x) in Eq. 8-6.We then have

or (8-10)

To associate a potential energy value Uwith the block at position x,we

choose the reference configuration to be when the spring is at its relaxed length

and the block is at xi0. Then the elastic potential energy Uiis 0, and Eq. 8-10

U1

2kxf

21

2kxi

U

xf

(kx)dx kxf

xdx1





The gravitational potential energy associated with a particle– Earth system

depends only on the vertical position y(or height) of the particle relative to the

reference position y0, not on the horizontal position.

183

8-1 POTENTIAL ENERGY

Checkpoint 2

A particle is to move along an xaxis from x0 to x1while a conser-

vative force, directed along the xaxis, acts on the particle.The figure

shows three situations in which the xcomponent of that force varies

with x.The force has the same maximum magnitude F1in all three sit-

uations.Rank the situations according to the change in the associated

potential energy during the particle’s motion,most positive first.

F1F1

–F1

(

)(

)

the ground, (3) at the limb, and (4) 1.0 m above the limb?

Take the gravitational potential energy to be zero at y0.

KEY IDEA

Once we have chosen the reference point for y0, we can

calculate the gravitational potential energy Uof the system

relative to that reference point with Eq. 8-9.

Calculations: For choice (1) the sloth is at y5.0 m, and

Umgy (2.0 kg)(9.8 m/s2)(5.0 m)

98 J. (Answer)

For the other choices, the values of Uare

(2) Umgy mg(2.0 m) 39 J,

(3) Umgy mg(0) 0J,

(4) Umgy mg(1.0 m)

19.6 J 20 J. (Answer)

(b) The sloth drops to the ground. For each choice of refer-

ence point, what is the change Uin the potential energy of

the sloth–Earth system due to the fall?

KEY IDEA

The change in potential energy does not depend on the

choice of the reference point for y0; instead, it depends

on the change in height y.

Calculation: For all four situations, we have the same y

5.0 m.Thus,for (1) to (4), Eq. 8-7 tells us that

Umg y(2.0 kg)(9.8 m/s2)(5.0 m)

98 J. (Answer)

Sample Problem 8.02 Choosing reference level for gravitational potential energy, sloth

Here is an example with this lesson plan: Generally you can

choose any level to be the reference level, but once chosen,

be consistent. A 2.0 kg sloth hangs 5.0 m above the ground

(Fig. 8-6).

(a) What is the gravitational potential energy Uof the

sloth–Earth system if we take the reference point y0 to be

(1) at the ground, (2) at a balcony floor that is 3.0 m above

Additional examples, video, and practice available at WileyPLUS

0 –3 –5 –6

3 0 –2 –3

5 2 0

6 3 1 0

(1) (2) (3) (4)

Figure 8-6 Four choices of reference point y0. Each yaxis is marked

in units of meters.The choice affects the value of the potential energy

Uof the sloth–Earth system. However, it does not affect the change

Uin potential energy of the system if the sloth moves by, say, falling.

becomes

which gives us

(elastic potential energy). (8-11)U(x)1

2kx2

U01

2kx20,

Conservation of Mechanical Energy

The mechanical energy Emec of a system is the sum of its potential energy Uand

the kinetic energy Kof the objects within it:

Emec KU(mechanical energy). (8-12)

In this module, we examine what happens to this mechanical energy when only

conservative forces cause energy transfers within the system—that is, when

frictional and drag forces do not act on the objects in the system. Also, we shall

assume that the system is isolated from its environment; that is, no external force

from an object outside the system causes energy changes inside the system.

When a conservative force does work Won an object within the system, that

force transfers energy between kinetic energy Kof the object and potential

energy Uof the system. From Eq. 7-10, the change Kin kinetic energy is

KW(8-13)

and from Eq. 8-1, the change Uin potential energy is

UW. (8-14)

Combining Eqs. 8-13 and 8-14, we find that

KU. (8-15)

In words, one of these energies increases exactly as much as the other decreases.

We can rewrite Eq. 8-15 as

K2K1(U2U1), (8-16)

where the subscripts refer to two different instants and thus to two different

arrangements of the objects in the system. Rearranging Eq. 8-16 yields

K2U2K1U1(conservation of mechanical energy). (8-17)

In words, this equation says:



the sum of K and U for

any state of a system







the sum of K and U for

any other state of the system



184 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

8-2 CONSERVATION OF MECHANICAL ENERGY

After reading this module, you should be able to . . .

8.05 After first clearly defining which objects form a system,

identify that the mechanical energy of the system is the

sum of the kinetic energies and potential energies of those

objects.

8.06 For an isolated system in which only conservative forces

act, apply the conservation of mechanical energy to relate

the initial potential and kinetic energies to the potential and

kinetic energies at a later instant.

Learning Objectives

Key Ideas

●The mechanical energy Emec of a system is the sum of its

kinetic energy Kand potential energy U:

Emec KU.

●An isolated system is one in which no external force causes

energy changes. If only conservative forces do work within

an isolated system, then the mechanical energy Emec of the

system cannot change. This principle of conservation of

mechanical energy is written as

K2U2K1U1,

in which the subscripts refer to different instants during an

energy transfer process. This conservation principle can also

be written as

Emec KU0.

©AP/Wide World Photos

In olden days, a person would be tossed

via a blanket to be able to see farther

over the flat terrain. Nowadays, it is

done just for fun. During the ascent of

the person in the photograph, energy is

transferred from kinetic energy to gravita-

tional potential energy. The maximum

height is reached when that transfer is

complete. Then the transfer is reversed

during the fall.

when the system is isolated and only conservative forces act on the objects in the

system. In other words:

185

8-2 CONSERVATION OF MECHANICAL ENERGY

In an isolated system where only conservative forces cause energy changes, the

kinetic energy and potential energy can change, but their sum, the mechanical

energy Emec of the system, cannot change.

When the mechanical energy of a system is conserved, we can relate the sum of kinetic

energy and potential energy at one instant to that at another instant without consider-

ing the intermediate motion and without finding the work done by the forces involved.

This result is called the principle of conservation of mechanical energy. (Now you

can see where conservative forces got their name.) With the aid of Eq. 8-15, we

can write this principle in one more form, as

Emec KU0. (8-18)

The principle of conservation of mechanical energy allows us to solve

problems that would be quite difficult to solve using only Newton’s laws:

Figure 8-7 shows an example in which the principle of conservation of

mechanical energy can be applied: As a pendulum swings, the energy of the

Figure 8-7 A pendulum, with its mass

concentrated in a bob at the lower end,

swings back and forth. One full cycle of

the motion is shown. During the cycle the

values of the potential and kinetic ener-

gies of the pendulum–Earth system vary

as the bob rises and falls, but the mechani-

cal energy Emec of the system remains

constant. The energy Emec can be

described as continuously shifting between

the kinetic and potential forms. In stages

(a) and (e), all the energy is kinetic energy.

The bob then has its greatest speed and is

at its lowest point. In stages (c) and (g), all

the energy is potential energy. The bob

then has zero speed and is at its highest

point. In stages (b), (d), ( f), and (h), half

the energy is kinetic energy and half is

potential energy. If the swinging involved

a frictional force at the point where the

pendulum is attached to the ceiling, or a

drag force due to the air, then Emec would

not be conserved, and eventually the

pendulum would stop.

(a)

(b)

(c)

(d)

(e)

(h)

(f)

(g)

v = +vmax

v = 0

v = –vmax

v = 0

All potential

energy

All potential

energy

The total energy

does not change

(it is conserved).

All kinetic energy

pendulum–Earth system is transferred back and forth between kinetic energy K

and gravitational potential energy U, with the sum KUbeing constant. If we

know the gravitational potential energy when the pendulum bob is at its highest

point (Fig. 8-7c), Eq. 8-17 gives us the kinetic energy of the bob at the lowest

point (Fig.8-7e).

For example, let us choose the lowest point as the reference point, with the

gravitational potential energy U20. Suppose then that the potential energy at

the highest point is U120 J relative to the reference point. Because the bob mo-

mentarily stops at its highest point, the kinetic energy there is K10. Putting these

values into Eq. 8-17 gives us the kinetic energy K2at the lowest point:

K20020 J or K220 J.

Note that we get this result without considering the motion between the highest

and lowest points (such as in Fig. 8-7d) and without finding the work done by any

forces involved in the motion.

186 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Checkpoint 3

The figure shows four

situations—one in

which an initially sta-

tionary block is dropped

and three in which the

block is allowed to slide

down frictionless ramps.

(a) Rank the situations

according to the kinetic energy of the block at point B,greatest first.(b) Rank them

according to the speed of the block at point B,greatest first.

B B B B

(

)(

)

System: Because the only force doing work on the child

is the gravitational force, we choose the child–Earth system

as our system, which we can take to be isolated.

Thus, we have only a conservative force doing work in

an isolated system, so we can use the principle of conserva-

tion of mechanical energy.

Calculations: Let the mechanical energy be Emec,twhen the

child is at the top of the slide and Emec,bwhen she is at the

bottom.Then the conservation principle tells us

Emec,bEmec,t. (8-19)

Sample Problem 8.03 Conservation of mechanical energy, water slide

The huge advantage of using the conservation of energy in-

stead of Newton’s laws of motion is that we can jump from

the initial state to the final state without considering all the

intermediate motion. Here is an example. In Fig. 8-8, a child

of mass mis released from rest at the top of a water slide,

at height h8.5 m above the bottom of the slide.

Assuming that the slide is frictionless because of the water

on it, find the child’s speed at the bottom of the slide.

KEY IDEAS

(1) We cannot find her speed at the bottom by using her ac-

celeration along the slide as we might have in earlier chap-

ters because we do not know the slope (angle) of the slide.

However, because that speed is related to her kinetic en-

ergy, perhaps we can use the principle of conservation of

mechanical energy to get the speed. Then we would not

need to know the slope. (2) Mechanical energy is conserved

in a system if the system is isolated and if only conservative

forces cause energy transfers within it. Let’s check.

Forces: Two forces act on the child. The gravitational

force, a conservative force, does work on her. The normal

force on her from the slide does no work because its direc-

tion at any point during the descent is always perpendicular

to the direction in which the child moves.

Figure 8-8 A child slides down a water slide as she descends a

height h.

The total mechanical

energy at the top

is equal to the total

at the bottom.

Reading a Potential Energy Curve

Once again we consider a particle that is part of a system in which a conservative

force acts. This time suppose that the particle is constrained to move along an

xaxis while the conservative force does work on it.We want to plot the potential

energy U(x) that is associated with that force and the work that it does, and then

we want to consider how we can relate the plot back to the force and to the kinetic

energy of the particle. However, before we discuss such plots, we need one more

relationship between the force and the potential energy.

Finding the Force Analytically

Equation 8-6 tells us how to find the change Uin potential energy between two

points in a one-dimensional situation if we know the force F(x). Now we want to

187

8-3 READING A POTENTIAL ENERGY CURVE

To show both kinds of mechanical energy, we have

KbUbKtUt, (8-20)

Dividing by mand rearranging yield

Putting vt0 and ytybhleads to

(Answer)13 m/s.

vb12gh 1(2)(9.8 m/s2)(8.5 m)

2vt

22g(ytyb).

2mvb

2mgyb1

2mvt

2mgyt.

Additional examples, video, and practice available at WileyPLUS

8-3 READING A POTENTIAL ENERGY CURVE

After reading this module, you should be able to . . .

8.07 Given a particle’s potential energy as a function of its

position x, determine the force on the particle.

8.08 Given a graph of potential energy versus x, determine

the force on a particle.

8.09 On a graph of potential energy versus x, superimpose a

line for a particle’s mechanical energy and determine the

particle’s kinetic energy for any given value of x.

8.10 If a particle moves along an xaxis, use a potential-

energy graph for that axis and the conservation of mechan-

ical energy to relate the energy values at one position to

those at another position.

8.11 On a potential-energy graph, identify any turning points

and any regions where the particle is not allowed because

of energy requirements.

8.12 Explain neutral equilibrium, stable equilibrium, and

unstable equilibrium.

Learning Objectives

Key Ideas

●If we know the potential energy function U(x)for a system

in which a one-dimensional force F(x)acts on a particle, we

can find the force as

●If U(x)is given on a graph, then at any value of x, the force

F(x)is the negative of the slope of the curve there and the

F(x) dU(x)

dx .

kinetic energy of the particle is given by

K(x)Emec U(x),

where Emec is the mechanical energy of the system.

●A turning point is a point xat which the particle reverses its

motion (there, K0).

●The particle is in equilibrium at points where the slope of

the U(x)curve is zero (there, F(x)0).

This is the same speed that the child would reach if she fell

8.5 m vertically. On an actual slide, some frictional forces

would act and the child would not be moving quite so fast.

Comments: Although this problem is hard to solve directly

with Newton’s laws, using conservation of mechanical en-

ergy makes the solution much easier. However, if we were

asked to find the time taken for the child to reach the bot-

tom of the slide, energy methods would be of no use; we

would need to know the shape of the slide, and we would

have a difficult problem.

go the other way; that is, we know the potential energy function U(x) and want

to find the force.

For one-dimensional motion, the work Wdone by a force that acts on a parti-

cle as the particle moves through a distance xis F(x)x. We can then write

Eq. 8-1 as

U(x)WF(x)x. (8-21)

Solving for F(x) and passing to the differential limit yield

(one-dimensional motion), (8-22)

which is the relation we sought.

We can check this result by putting , which is the elastic poten-

tial energy function for a spring force. Equation 8-22 then yields, as expected,

F(x)kx, which is Hooke’s law. Similarly, we can substitute U(x)mgx,

which is the gravitational potential energy function for a particle–Earth system,

with a particle of mass mat height xabove Earth’s surface. Equation 8-22 then

yields Fmg, which is the gravitational force on the particle.

The Potential Energy Curve

Figure 8-9ais a plot of a potential energy function U(x) for a system in which a

particle is in one-dimensional motion while a conservative force F(x) does work

on it.We can easily find F(x) by (graphically) taking the slope of the U(x) curve at

various points. (Equation 8-22 tells us that F(x) is the negative of the slope of the

U(x) curve.) Figure 8-9bis a plot of F(x) found in this way.

Turning Points

In the absence of a nonconservative force, the mechanical energy Eof a system

has a constant value given by

U(x)K(x)Emec. (8-23)

Here K(x) is the kinetic energy function of a particle in the system (this K(x)

gives the kinetic energy as a function of the particle’s location x). We may

rewrite Eq. 8-23 as

K(x)Emec U(x). (8-24)

Suppose that Emec (which has a constant value, remember) happens to be 5.0 J. It

would be represented in Fig. 8-9cby a horizontal line that runs through the value

5.0 J on the energy axis. (It is, in fact, shown there.)

Equation 8-24 and Fig. 8-9dtell us how to determine the kinetic energy Kfor

any location xof the particle: On the U(x) curve, find Ufor that location xand

then subtract Ufrom Emec. In Fig. 8-9efor example, if the particle is at any point

to the right of x5, then K1.0 J.The value of Kis greatest (5.0 J) when the parti-

cle is at x2and least (0 J) when the particle is at x1.

Since Kcan never be negative (because v2is always positive), the particle can

never move to the left of x1, where Emec Uis negative. Instead, as the particle

moves toward x1from x2,Kdecreases (the particle slows) until K0 at x1(the

particle stops there).

Note that when the particle reaches x1, the force on the particle, given by

Eq. 8-22, is positive (because the slope dU/dx is negative). This means that the

particle does not remain at x1but instead begins to move to the right, opposite its

earlier motion. Hence x1is a turning point, a place where K0 (because UE)

and the particle changes direction. There is no turning point (where K0) on

the right side of the graph. When the particle heads to the right, it will continue

indefinitely.

U(x)1

2kx2

F(x)dU(x)

188 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

189

8-3 READING A POTENTIAL ENERGY CURVE

U(J)

x1x3x4x5

(a)

U(x)

–

(b)

x1x3x4x5

F (N)

Mild force, –x direction

Strong force, +xdirection

This is a plot of the potential

energy U versus position x.

Force is equal to the negative of

the slope of the U(x) plot.

U (J), Emec (J)

x1x3x4x5

(c)

Emec = 5.0 J

U(x)

The flat line shows a given value of

the total mechanical energy Emec.

The difference between the total energy

and the potential energy is the

kinetic energy K.

U (J), Emec (J)

x1x3x4x5

(d)

Emec = 5.0 J

U(x)

(f)x2

x1x3x4x5

U (J), Emec (J)

At this position, K is greatest and

the particle is moving the fastest.

At this position, K is zero (a turning point).

The particle cannot go farther to the left.

For either of these three choices for Emec,

the particle is trapped (cannot escape

left or right).

U (J), Emec (J)

x1x3x4x5

(e)

Emec = 5.0 J

K = 1.0 J at x > x5

K = 5.0 J at x2

Figure 8-9 (a) A plot of U(x), the potential energy function of a system containing a particle confined to move along an xaxis. There is no

friction, so mechanical energy is conserved. (b) A plot of the force F(x) acting on the particle, derived from the potential energy plot by

taking its slope at various points. (c)–(e) How to determine the kinetic energy. ( f) The U(x) plot of (a) with three possible values of Emec

shown. In WileyPLUS, this figure is available as an animation with voiceover.

Equilibrium Points

Figure 8-9fshows three different values for Emec superposed on the plot of the

potential energy function U(x) of Fig. 8-9a. Let us see how they change the situa-

tion. If Emec 4.0 J (purple line), the turning point shifts from x1to a point

between x1and x2. Also, at any point to the right of x5, the system’s mechanical

energy is equal to its potential energy; thus, the particle has no kinetic energy and

(by Eq. 8-22) no force acts on it, and so it must be stationary. A particle at such a

position is said to be in neutral equilibrium. (A marble placed on a horizontal

tabletop is in that state.)

If Emec 3.0 J (pink line), there are two turning points: One is between

x1and x2, and the other is between x4and x5. In addition, x3is a point at which

K0. If the particle is located exactly there, the force on it is also zero, and the

particle remains stationary. However, if it is displaced even slightly in either

direction, a nonzero force pushes it farther in the same direction, and the particle

continues to move. A particle at such a position is said to be in unstable equilib-

rium. (A marble balanced on top of a bowling ball is an example.)

Next consider the particle’s behavior if Emec 1.0 J (green line). If we place it

at x4,it is stuck there. It cannot move left or right on its own because to do so would

require a negative kinetic energy. If we push it slightly left or right, a restoring force

appears that moves it back to x4.A particle at such a position is said to be in stable

equilibrium. (A marble placed at the bottom of a hemispherical bowl is an example.)

If we place the particle in the cup-like potential well centered at x2,it is between two

turning points. It can still move somewhat, but only partway to x1or x3.

190 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Checkpoint 4

The figure gives the potential energy function

U(x) for a system in which a particle is in one-

dimensional motion. (a) Rank regions AB,BC, and

CD according to the magnitude of the force on the

particle, greatest first. (b) What is the direction of

the force when the particle is in region AB?

U(x) (J)

A B C D x

Calculations: At ,the particle has kinetic energy

Because the potential energy there is , the mechanical

energy is

This value for is plotted as a horizontal line in Fig. 8-10a.

From that figure we see that at , the potential

energy is . The kinetic energy is the difference

between and :

Because , we find

. (Answer)

(b) Where is the particle’s turning point located?

v13.0 m/s

K11

2mv1

K1Emec U116.0 J 7.0 J 9.0 J

Emec

U17.0 J

x4.5 m

Emec

Emec K0U016.0 J 016.0 J

U0

16.0 J.

K01

2mv2

01

2(2.00 kg)(4.00 m/s)2

x6.5 m

Sample Problem 8.04 Reading a potential energy graph

A 2.00 kg particle moves along an xaxis in one-dimensional

motion while a conservative force along that axis acts on it.

The potential energy U(x) associated with the force is plot-

ted in Fig. 8-10a. That is, if the particle were placed at any

position between and , it would have the

plotted value of U. At , the particle has velocity

(a) From Fig. 8-10a, determine the particle’s speed at

KEY IDEAS

(1) The particle’s kinetic energy is given by Eq. 7-1

( ). (2) Because only a conservative force acts on

the particle, the mechanical energy is con-

served as the particle moves. (3) Therefore, on a plot of U(x)

such as Fig. 8-10a, the kinetic energy is equal to the differ-

ence between and U.Emec

Emec (KU)

K1

2mv2

x14.5 m

:(4.00 m/s)

x6.5 m

x7.00 mx0

Additional examples, video, and practice available at WileyPLUS

191

8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE

KEY IDEA

The turning point is where the force momentarily stops and

then reverses the particle’s motion. That is, it is where the

particle momentarily has and thus .

Calculations: Because Kis the difference between

, we want the point in Fig. 8-10awhere the plot of

Urises to meet the horizontal line of , as shown in Fig.

8-10b. Because the plot of Uis a straight line in Fig. 8-10b,

we can draw nested right triangles as shown and then write

the proportionality of distances

which gives us .Thus, the turning point is at

. (Answer)

region .

KEY IDEA

The force is given by Eq. 8-22 (F(x)dU(x)/dx):The force

is equal to the negative of the slope on a graph of U(x).

Calculations: For the graph of Fig. 8-10b, we see that for

the range the force is

. (Answer)F 20 J 7.0 J

1.0 m 4.0 m 4.3 N

1.0 m x4.0 m

1.9 m x4.0 m

x4.0 m d1.9 m

d2.08 m

16 7.0

d20 7.0

4.0 1.0

Emec

Emec and U

K0v0

Thus, the force has magnitude 4.3 N and is in the positive

direction of the xaxis. This result is consistent with the fact

that the initially leftward-moving particle is stopped by the

force and then sent rightward.

Figure 8-10 (a) A plot of potential energy Uversus position x.

(b) A section of the plot used to find where the particle turns

around.

Kinetic energy is the difference

between the total energy and

the potential energy.

Emec = 16 J

01 4567

x (m)

(a)

U ( J)

Turning point

714

x (m)

(b)

U ( J)

The kinetic energy is zero

at the turning point (the

particle speed is zero).

8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE

After reading this module, you should be able to . . .

8.13 When work is done on a system by an external force

with no friction involved, determine the changes in kinetic

energy and potential energy.

8.14 When work is done on a system by an external force

with friction involved, relate that work to the changes in

kinetic energy, potential energy, and thermal energy.

●Work Wis energy transferred to or from a system by means

of an external force acting on the system.

●When more than one force acts on a system, their net work

is the transferred energy.

●When friction is not involved, the work done on the system

and the change Emec in the mechanical energy of the system

are equal:

WEmec KU.

●When a kinetic frictional force acts within the system, then

the thermal energy Eth of the system changes. (This energy is

associated with the random motion of atoms and molecules

in the system.) The work done on the system is then

WEmec Eth.

●The change Eth is related to the magnitude fkof the frictional

force and the magnitude dof the displacement caused by the

external force by

Eth fkd.

Learning Objectives

Key Ideas

Work Done on a System by an External Force

In Chapter 7, we defined work as being energy transferred to or from an object

by means of a force acting on the object.We can now extend that definition to an

external force acting on a system of objects.

192 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Work is energy transferred to or from a system by means of an external force

acting on that system.

Figure 8-11arepresents positive work (a transfer of energy to a system), and

Fig. 8-11brepresents negative work (a transfer of energy from a system). When

more than one force acts on a system, their net work is the energy transferred to

or from the system.

These transfers are like transfers of money to and from a bank account. If a

system consists of a single particle or particle-like object, as in Chapter 7, the

work done on the system by a force can change only the kinetic energy of the

system.The energy statement for such transfers is the work–kinetic energy theo-

rem of Eq. 7-10 (KW); that is, a single particle has only one energy account,

called kinetic energy. External forces can transfer energy into or out of that

account. If a system is more complicated, however, an external force can change

other forms of energy (such as potential energy); that is, a more complicated

system can have multiple energy accounts.

Let us find energy statements for such systems by examining two basic situa-

tions, one that does not involve friction and one that does.

No Friction Involved

To compete in a bowling-ball-hurling contest, you first squat and cup your hands

under the ball on the floor.Then you rapidly straighten up while also pulling your

hands up sharply, launching the ball upward at about face level. During your

upward motion, your applied force on the ball obviously does work; that is, it is an

external force that transfers energy, but to what system?

To answer, we check to see which energies change. There is a change Kin

the ball’s kinetic energy and, because the ball and Earth become more sepa-

rated, there is a change Uin the gravitational potential energy of the

ball–Earth system.To include both changes, we need to consider the ball–Earth

system. Then your force is an external force doing work on that system, and the

work is

WKU, (8-25)

or WEmec (work done on system, no friction involved), (8-26)

where Emec is the change in the mechanical energy of the system. These two

equations, which are represented in Fig. 8-12, are equivalent energy statements

for work done on a system by an external force when friction is not involved.

Friction Involved

We next consider the example in Fig. 8-13a. A constant horizontal force pulls a

block along an xaxis and through a displacement of magnitude d, increasing the

block’s velocity from to . During the motion, a constant kinetic frictional

force from the floor acts on the block. Let us first choose the block as our

system and apply Newton’s second law to it. We can write that law for compo-

nents along the xaxis (Fnet,xmax) as

Ffkma. (8-27)

Positive W

ystem

(a)

Negative W

System

(

)

Figure 8-11 (a) Positive work Wdone on an

arbitrary system means a transfer of

energy to the system. (b) Negative work

Wmeans a transfer of energy from the

system.

WΔEmec =ΔK+ΔU

Ball–Earth

system

Your lifting force

transfers energy to

kinetic energy and

potential energy.

Figure 8-12 Positive work Wis done on a

system of a bowling ball and Earth, caus-

ing a change Emec in the mechanical

energy of the system, a change Kin the

ball’s kinetic energy, and a change Uin

the system’s gravitational potential energy.

Because the forces are constant, the acceleration is also constant. Thus, we can

use Eq. 2-16 to write

Solving this equation for a, substituting the result into Eq. 8-27, and rearranging

then give us

(8-28)

or,because for the block,

Fd Kfkd. (8-29)

In a more general situation (say, one in which the block is moving up a ramp), there

can be a change in potential energy. To include such a possible change, we general-

ize Eq. 8-29 by writing

Fd Emec fkd. (8-30)

By experiment we find that the block and the portion of the floor along

which it slides become warmer as the block slides. As we shall discuss in

Chapter 18, the temperature of an object is related to the object’s thermal energy

Eth (the energy associated with the random motion of the atoms and molecules in

the object). Here, the thermal energy of the block and floor increases because

(1) there is friction between them and (2) there is sliding. Recall that friction is

due to the cold-welding between two surfaces. As the block slides over the floor,

the sliding causes repeated tearing and re-forming of the welds between the

block and the floor, which makes the block and floor warmer. Thus, the sliding

increases their thermal energy Eth.

Through experiment, we find that the increase Eth in thermal energy is

equal to the product of the magnitudes fkand d:

Eth fkd(increase in thermal energy by sliding). (8-31)

Thus, we can rewrite Eq. 8-30 as

Fd Emec Eth. (8-32)

Fd is the work Wdone by the external force (the energy transferred by the

force), but on which system is the work done (where are the energy transfers made)?

To answer, we check to see which energies change. The block’s mechanical energy

2mv21

2mv0

2K

Fd 1

2mv21

2mv0

2fkd

v2v0

22ad.

193

8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE

v0v

(a)

The applied force supplies energy.

The frictional force transfers some

of it to thermal energy.

(b)

Block–floor

system

ΔEmec

ΔEth

So, the work done by the applied

force goes into kinetic energy

and also thermal energy.

Figure 8-13 (a) A block is pulled across a floor by force while a kinetic frictional

force opposes the motion. The block has velocity at the start of a displacement

and velocity at the end of the displacement. (b) Positive work Wis done on the

block–floor system by force , resulting in a change Emec in the block’s mechanical

energy and a change Eth in the thermal energy of the block and floor.

194 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Checkpoint 5

In three trials,a block is pushed

by a horizontal applied force

across a floor that is not friction-

less,as in Fig. 8-13a.The magni-

tudes Fof the applied force and

the results of the pushing on the

block’s speed are given in the

table. In all three trials, the block is pushed through the same distance d. Rank the

three trials according to the change in the thermal energy of the block and floor that

occurs in that distance d, greatest first.

Trial FResult on Block’s Speed

a 5.0 N decreases

b 7.0 N remains constant

c 8.0 N increases

be friction and a change Eth in thermal energy of the crate

and the floor. Therefore, the system on which the work is

done is the crate–floor system, because both energy

changes occur in that system.

(b) What is the increase Eth in the thermal energy of the

crate and floor?

KEY IDEA

We can relate Eth to the work Wdone by with the energy

statement of Eq. 8-33 for a system that involves friction:

WEmec Eth. (8-34)

Calculations: We know the value of Wfrom (a). The

change Emec in the crate’s mechanical energy is just the

change in its kinetic energy because no potential energy

changes occur,so we have

Substituting this into Eq.8-34 and solving for Eth,we find

(Answer)

Without further experiments, we cannot say how much of

this thermal energy ends up in the crate and how much in

the floor.We simply know the total amount.

22.2 J 22 J.

20 J 1

2(14 kg)[(0.20 m/s)2(0.60 m/s)2]

Eth W(1

2mv21

2mv0

2)W1

2m(v2v0

Emec K1

2mv21

2mv0

Sample Problem 8.05 Work, friction, change in thermal energy, cabbage heads

A food shipper pushes a wood crate of cabbage heads (total

mass m14 kg) across a concrete floor with a constant

horizontal force of magnitude 40 N. In a straight-line dis-

placement of magnitude d0.50 m, the speed of the crate

decreases from v00.60 m/s to v0.20 m/s.

(a) How much work is done by force , and on what system

does it do the work?

KEY IDEA

Because the applied force is constant, we can calculate

the work it does by using Eq. 7-7 ( ).

Calculation: Substituting given data, including the fact that

force and displacement are in the same direction, we

find

WFd cos f(40 N)(0.50 m) cos 0

20 J. (Answer)

Reasoning: To determine the system on which the work is

done, let’s check which energies change. Because the crate’s

speed changes, there is certainly a change Kin the crate’s

kinetic energy. Is there friction between the floor and the

crate, and thus a change in thermal energy? Note that and

the crate’s velocity have the same direction.Thus, if there is

no friction, then should be accelerating the crate to a

greater speed. However, the crate is slowing, so there must

WFd cos







Additional examples, video, and practice available at WileyPLUS

changes, and the thermal energies of the block and floor also change. Therefore, the

work done by force is done on the block–floor system.That work is

WEmec Eth (work done on system, friction involved). (8-33)

This equation, which is represented in Fig. 8-13b, is the energy statement for the

work done on a system by an external force when friction is involved.

Conservation of Energy

We now have discussed several situations in which energy is transferred to or

from objects and systems, much like money is transferred between accounts.

In each situation we assume that the energy that was involved could always be

accounted for; that is, energy could not magically appear or disappear. In more

formal language, we assumed (correctly) that energy obeys a law called the law of

conservation of energy, which is concerned with the total energy Eof a system.

That total is the sum of the system’s mechanical energy, thermal energy, and any

type of internal energy in addition to thermal energy. (We have not yet discussed

other types of internal energy.) The law states that

195

8-5 CONSERVATION OF ENERGY

The total energy Eof a system can change only by amounts of energy that are

transferred to or from the system.

8-5 CONSERVATION OF ENERGY

After reading this module, you should be able to . . .

8.15 For an isolated system (no net external force), apply the

conservation of energy to relate the initial total energy

(energies of all kinds) to the total energy at a later instant.

8.16 For a nonisolated system, relate the work done on the

system by a net external force to the changes in the vari-

ous types of energies within the system.

8.17 Apply the relationship between average power, the

associated energy transfer, and the time interval in which

that transfer is made.

8.18 Given an energy transfer as a function of time (either as

an equation or a graph), determine the instantaneous

power (the transfer at any given instant).

Learning Objectives

●The total energy Eof a system (the sum of its mechanical

energy and its internal energies, including thermal energy)

can change only by amounts of energy that are transferred to

or from the system. This experimental fact is known as the law

of conservation of energy.

●If work Wis done on the system, then

WEEmec Eth Eint.

If the system is isolated (W0), this gives

Emec Eth Eint 0

and Emec,2 Emec,1 Eth Eint,

where the subscripts 1 and 2 refer to two different instants.

●The power due to a force is the rate at which that force

transfers energy. If an amount of energy Eis transferred by

a force in an amount of time t, the average power of the

force is

●The instantaneous power due to a force is

On a graph of energy Eversus time t, the power is the slope

of the plot at any given time.

PdE

dt .

Pavg E

t.

Key Ideas

The only type of energy transfer that we have considered is work Wdone on a

system by an external force.Thus, for us at this point, this law states that

WEEmec Eth Eint, (8-35)

where Emec is any change in the mechanical energy of the system, Eth is any

change in the thermal energy of the system, and Eint is any change in any

other type of internal energy of the system. Included in Emec are changes Kin

kinetic energy and changes Uin potential energy (elastic, gravitational, or any

other type we might find).

This law of conservation of energy is not something we have derived from

basic physics principles. Rather, it is a law based on countless experiments.

Scientists and engineers have never found an exception to it. Energy simply can-

not magically appear or disappear.

Isolated System

If a system is isolated from its environment, there can be no energy transfers to or

from it. For that case,the law of conservation of energy states:

196 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

The total energy Eof an isolated system cannot change.

Many energy transfers may be going on within an isolated system—between,

say, kinetic energy and a potential energy or between kinetic energy and ther-

mal energy. However, the total of all the types of energy in the system cannot

change. Here again, energy cannot magically appear or disappear.

We can use the rock climber in Fig. 8-14 as an example, approximating

him, his gear, and Earth as an isolated system. As he rappels down the rock

face, changing the configuration of the system, he needs to control the transfer

of energy from the gravitational potential energy of the system. (That energy

cannot just disappear.) Some of it is transferred to his kinetic energy.

However, he obviously does not want very much transferred to that type or he

will be moving too quickly, so he has wrapped the rope around metal rings to

produce friction between the rope and the rings as he moves down.The sliding

of the rings on the rope then transfers the gravitational potential energy of the

system to thermal energy of the rings and rope in a way that he can control.

The total energy of the climber–gear–Earth system (the total of its gravita-

tional potential energy, kinetic energy, and thermal energy) does not change

during his descent.

For an isolated system, the law of conservation of energy can be written in

two ways. First, by setting W0 in Eq. 8-35, we get

Emec Eth Eint 0(isolated system). (8-36)

We can also let Emec Emec,2 Emec,1, where the subscripts 1 and 2 refer to two

different instants—say, before and after a certain process has occurred.Then Eq.

8-36 becomes

Emec,2 Emec,1 Eth Eint. (8-37)

Equation 8-37 tells us:

Figure 8-14 To descend, the rock climber

must transfer energy from the gravitational

potential energy of a system consisting of

him, his gear, and Earth. He has wrapped

the rope around metal rings so that the

rope rubs against the rings. This allows

most of the transferred energy to go to the

thermal energy of the rope and rings

rather than to his kinetic energy.

Tyler Stableford/The Image Bank/Getty Images

In an isolated system, we can relate the total energy at one instant to the total

energy at another instant without considering the energies at intermediate times.

This fact can be a very powerful tool in solving problems about isolated systems

when you need to relate energies of a system before and after a certain process

occurs in the system.

In Module 8-2, we discussed a special situation for isolated systems—namely,

the situation in which nonconservative forces (such as a kinetic frictional force)

do not act within them. In that special situation, Eth and Eint are both zero, and

so Eq. 8-37 reduces to Eq. 8-18. In other words, the mechanical energy of an

isolated system is conserved when nonconservative forces do not act in it.

External Forces and Internal Energy Transfers

An external force can change the kinetic energy or potential energy of an object

without doing work on the object—that is, without transferring energy to the

object. Instead, the force is responsible for transfers of energy from one type to

another inside the object.

Figure 8-15 shows an example.An initially stationary ice-skater pushes away

from a railing and then slides over the ice (Figs. 8-15aand b). Her kinetic energy

increases because of an external force on her from the rail. However, that force

does not transfer energy from the rail to her. Thus, the force does no work on

her. Rather, her kinetic energy increases as a result of internal transfers from the

biochemical energy in her muscles.

Figure 8-16 shows another example. An engine increases the speed of a car

with four-wheel drive (all four wheels are made to turn by the engine). During

the acceleration, the engine causes the tires to push backward on the road sur-

face. This push produces frictional forces that act on each tire in the forward

direction. The net external force from the road, which is the sum of these fric-

tional forces, accelerates the car, increasing its kinetic energy. However, does

not transfer energy from the road to the car and so does no work on the car.

Rather, the car’s kinetic energy increases as a result of internal transfers from the

energy stored in the fuel.

197

8-5 CONSERVATION OF ENERGY

Figure 8-15 (a) As a skater pushes herself away from a railing, the force on her from

the railing is . (b) After the skater leaves the railing, she has velocity . (c) External

force acts on the skater, at angle fwith a horizontal xaxis. When the skater goes

through displacement , her velocity is changed from (0) to by the horizontal

component of .F

Ice

(a)

(c)

(b)

Her push on the rail causes

a transfer of internal energy

to kinetic energy.

Figure 8-16 A vehicle accelerates to the

right using four-wheel drive. The road

exerts four frictional forces (two of them

shown) on the bottom surfaces of the tires.

Taken together, these four forces make up

the net external force acting on the car.F

acom

f f

In situations like these two, we can sometimes relate the external force on

an object to the change in the object’s mechanical energy if we can simplify the

situation. Consider the ice-skater example.During her push through distance din

Fig. 8-15c, we can simplify by assuming that the acceleration is constant, her

speed changing from v00 to v. (That is, we assume has constant magnitude F

and angle f.) After the push, we can simplify the skater as being a particle and

neglect the fact that the exertions of her muscles have increased the thermal

energy in her muscles and changed other physiological features. Then we can

apply Eq. 7-5 to write

KK0(Fcos f)d,

or KFd cos f. (8-38)

If the situation also involves a change in the elevation of an object, we can

include the change Uin gravitational potential energy by writing

UKFd cos f. (8-39)

The force on the right side of this equation does no work on the object but is still

responsible for the changes in energy shown on the left side.

Power

Now that you have seen how energy can be transferred from one type to another,

we can expand the definition of power given in Module 7-6. There power is

2mv21

2mv0

2Fxd)

defined as the rate at which work is done by a force. In a more general sense,

power Pis the rate at which energy is transferred by a force from one type to

another. If an amount of energy Eis transferred in an amount of time t, the

average power due to the force is

(8-40)

Similarly, the instantaneous power due to the force is

(8-41)PdE

dt .

avg E

t.

198 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

on the glider to get it moving, a spring force does work on

it, transferring energy from the elastic potential energy of

the compressed spring to kinetic energy of the glider. The

spring force also pushes against a rigid wall. Because there

is friction between the glider and the ground-level track,

the sliding of the glider along that track section increases

their thermal energies.

System: Let’s take the system to contain all the interact-

ing bodies: glider, track, spring, Earth, and wall. Then, be-

cause all the force interactions are within the system, the

system is isolated and thus its total energy cannot change.

So, the equation we should use is not that of some external

force doing work on the system. Rather, it is a conservation

of energy.We write this in the form of Eq. 8-37:

Emec,2 Emec,1 Eth. (8-42)

This is like a money equation: The final money is equal to

the initial money minus the amount stolen away by a thief.

Here, the final mechanical energy is equal to the initial me-

chanical energy minus the amount stolen away by friction.

None has magically appeared or disappeared.

Calculations: Now that we have an equation, let’s find

distance L. Let subscript 1 correspond to the initial state

of the glider (when it is still on the compressed spring)

and subscript 2 correspond to the final state of the glider

(when it has come to rest on the ground-level track). For

both states, the mechanical energy of the system is the

sum of any potential energy and any kinetic energy.

We have two types of potential energy: the elastic po-

tential energy (Uekx2) associated with the compressed

Sample Problem 8.06 Lots of energies at an amusement park water slide

Figure 8-17 shows a water-slide ride in which a glider is shot

by a spring along a water-drenched (frictionless) track that

takes the glider from a horizontal section down to ground

level.As the glider then moves along ground-level track, it is

gradually brought to rest by friction. The total mass of the

glider and its rider is m200 kg, the initial compression of

the spring is d5.00 m, the spring constant is k3.20 

103N/m, the initial height is h35.0 m, and the coefficient

of kinetic friction along the ground-level track is mk0.800.

Through what distance Ldoes the glider slide along the

ground-level track until it stops?

KEY IDEAS

Before we touch a calculator and start plugging numbers

into equations, we need to examine all the forces and then

determine what our system should be. Only then can we

decide what equation to write. Do we have an isolated sys-

tem (our equation would be for the conservation of en-

ergy) or a system on which an external force does work

(our equation would relate that work to the system’s

change in energy)?

Forces: The normal force on the glider from the track

does no work on the glider because the direction of this

force is always perpendicular to the direction of the

glider’s displacement. The gravitational force does work

on the glider, and because the force is conservative we can

associate a potential energy with it. As the spring pushes

spring and the gravitational potential energy (Ugmgy) as-

m⫽0

Figure 8-17 A spring-loaded amusement park water slide.

sociated with the glider’s elevation. For the latter, let’s take

ground level as the reference level. That means that the

glider is initially at height yhand finally at height y0.

In the initial state, with the glider stationary and ele-

vated and the spring compressed, the energy is

Emec,1  K1Ue1Ug1

 0kd2mgh. (8-43)

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199

REVIEW & SUMMARY

Substituting Eqs. 8-43 through 8-45 into Eq. 8-42, we find

0kd2mgh mkmgL, (8-46)

and

69.3 m. (Answer)

Finally, note how algebraically simple our solution is. By

carefully defining a system and realizing that we have an

isolated system, we get to use the law of the conservation of

energy. That means we can relate the initial and final states

of the system with no consideration of the intermediate

states. In particular,we did not need to consider the glider as

it slides over the uneven track. If we had, instead, applied

Newton’s second law to the motion, we would have had to

know the details of the track and would have faced a far

more difficult calculation.

(3.20 103 N/m)(5.00 m)2

2(0.800)(200 kg)(9.8 m/s2)35 m

0.800

kd2

2mkmg h

Conservative Forces A force is a conservative force if the net

work it does on a particle moving around any closed path, from an

initial point and then back to that point, is zero. Equivalently, a

force is conservative if the net work it does on a particle moving

between two points does not depend on the path taken by the par-

ticle. The gravitational force and the spring force are conservative

forces; the kinetic frictional force is a nonconservative force.

Potential Energy Apotential energy is energy that is associated

with the configuration of a system in which a conservative force acts.

When the conservative force does work Won a particle within the sys-

tem,the change Uin the potential energy of the system is

UW. (8-1)

If the particle moves from point xito point xf, the change in the

potential energy of the system is

(8-6)

Gravitational Potential Energy The potential energy asso-

ciated with a system consisting of Earth and a nearby particle is

gravitational potential energy. If the particle moves from height yi

to height yf, the change in the gravitational potential energy of the

particle–Earth system is

Umg(yfyi)mg y. (8-7)

If the reference point of the particle is set as yi0 and the cor-

responding gravitational potential energy of the system is set as

Ui0, then the gravitational potential energy Uwhen the parti-

U

xf

F(x)dx.

Review & Summary

cle is at any height yis

U(y)mgy. (8-9)

Elastic Potential Energy Elastic potential energy is the

energy associated with the state of compression or extension of an

elastic object. For a spring that exerts a spring force Fkx when

its free end has displacement x, the elastic potential energy is

(8-11)

The reference configuration has the spring at its relaxed length, at

which x0 and U0.

Mechanical Energy The mechanical energy Emec of a system

is the sum of its kinetic energy Kand potential energy U:

Emec KU. (8-12)

An isolated system is one in which no external force causes energy

changes. If only conservative forces do work within an isolated sys-

tem, then the mechanical energy Emec of the system cannot change.

This principle of conservation of mechanical energy is written as

K2U2K1U1, (8-17)

in which the subscripts refer to different instants during an

energy transfer process. This conservation principle can also be

written as

Emec KU0. (8-18)

Potential Energy Curves If we know the potential energy

function U(x) for a system in which a one-dimensional force F(x)

U(x)1

2kx2.

In the final state, with the spring now in its relaxed state and

the glider again stationary but no longer elevated, the final

mechanical energy of the system is

Emec,2  K2Ue2Ug2

 000. (8-44)

Let’s next go after the change Eth of the thermal energy of

the glider and ground-level track. From Eq. 8-31, we can

substitute for Eth with fkL(the product of the frictional

force magnitude and the distance of rubbing). From Eq. 6-2,

we know that fk mkFN, where FNis the normal force.

Because the glider moves horizontally through the region

with friction, the magnitude of FNis equal to mg (the up-

ward force matches the downward force). So, the friction’s

theft from the mechanical energy amounts to

Eth  mkmgL. (8-45)

(By the way, without further experiments, we cannot say

how much of this thermal energy ends up in the glider and

how much in the track. We simply know the total amount.)

200 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

acts on a particle, we can find the force as

(8-22)

If U(x) is given on a graph, then at any value of x, the force F(x) is

the negative of the slope of the curve there and the kinetic energy

of the particle is given by

K(x)Emec U(x), (8-24)

where Emec is the mechanical energy of the system.A turning point

is a point xat which the particle reverses its motion (there, K0).

The particle is in equilibrium at points where the slope of the U(x)

curve is zero (there, F(x)0).

Work Done on a System by an External Force Work W

is energy transferred to or from a system by means of an external

force acting on the system. When more than one force acts on a

system, their net work is the transferred energy. When friction is

not involved, the work done on the system and the change Emec in

the mechanical energy of the system are equal:

WEmec KU. (8-26, 8-25)

When a kinetic frictional force acts within the system, then the ther-

mal energy Eth of the system changes. (This energy is associated with

the random motion of atoms and molecules in the system.) The

work done on the system is then

WEmec Eth. (8-33)

F(x) dU(x)

dx .

The change Eth is related to the magnitude fkof the frictional force

and the magnitude dof the displacement caused by the external

force by

Eth fkd. (8-31)

Conservation of Energy The total energy Eof a system

(the sum of its mechanical energy and its internal energies,

including thermal energy) can change only by amounts of energy

that are transferred to or from the system. This experimental fact

is known as the law of conservation of energy. If work Wis done

on the system, then

WEEmec Eth Eint. (8-35)

If the system is isolated (W0), this gives

Emec Eth Eint 0 (8-36)

and Emec,2 Emec,1 Eth Eint, (8-37)

where the subscripts 1 and 2 refer to two different instants.

Power The power due to a force is the rate at which that force

transfers energy. If an amount of energy Eis transferred by

a force in an amount of time t, the average power of the force is

(8-40)

The instantaneous power due to a force is

(8-41)PdE

dt .

Pavg E

t.

1In Fig. 8-18, a horizontally moving block can take three fric-

tionless routes, differing only in elevation, to reach the dashed

finish line. Rank the routes according to (a) the speed of the block

at the finish line and (b) the travel time of the block to the finish

line, greatest first.

tude of the force on the particle, greatest first. What value must

the mechanical energy Emec of the particle not exceed if the par-

ticle is to be (b) trapped in the potential well at the left, (c)

trapped in the potential well at the right, and (d) able to move

between the two potential wells but not to the right of point H?

For the situation of (d), in which of regions BC, DE, and FG will

the particle have (e) the greatest kinetic energy and (f) the least

speed?

3Figure 8-20 shows one direct

path and four indirect paths from

point ito point f. Along the direct

path and three of the indirect paths,

only a conservative force Fcacts on

a certain object. Along the fourth

indirect path, both Fcand a noncon-

servative force Fnc act on the object.

The change Emec in the object’s

mechanical energy (in joules) in going from ito fis indicated along

each straight-line segment of the indirect paths. What is Emec (a)

from ito falong the direct path and (b) due to Fnc along the one

path where it acts?

4In Fig. 8-21, a small, initially stationary block is released on a

frictionless ramp at a height of 3.0 m. Hill heights along the ramp

are as shown in the figure.The hills have identical circular tops, and

the block does not fly off any hill. (a) Which hill is the first the

block cannot cross? (b) What does the block do after failing

to cross that hill? Of the hills that the block can cross, on which hill-

Questions

(1)

Finish line

(2)

(3)

Figure 8-18 Question 1.

A B

D E F

H x

U(x) (J)

Figure 8-19 Question 2.

–30

32 10

–10

–6 –4

15 7

–30

Figure 8-20 Question 3.

2Figure 8-19 gives the potential energy function of a particle.

(a) Rank regions AB, BC, CD, and DE according to the magni-

descends, it pulls on a block via a

second rope, and the block slides

over a lab table. Again consider the

cylinder–rod–Earth system, similar

to that shown in Fig. 8-23b. Your

work on the system is 200 J. The sys-

tem does work of 60 J on the block.

Within the system, the kinetic

energy increases by 130 J and

the gravitational potential energy

decreases by 20 J. (a) Draw an “en-

ergy statement” for the system, as in

Fig. 8-23c. (b) What is the change in

the thermal energy within the system?

8In Fig. 8-25, a block slides along a track that descends through

distance h. The track is frictionless except for the lower section.

There the block slides to a stop in a certain distance Dbecause of

friction. (a) If we decrease h, will the block now slide to a stop in a

distance that is greater than, less than, or equal to D? (b) If, instead,

we increase the mass of the block, will the stopping distance now be

greater than, less than, or equal to D?

201

QUESTIONS

top is (c) the centripetal acceleration of the block greatest and (d) the

normal force on the block least?

(

) (

)(

)

Figure 8-26 Question 9.

Figure 8-25 Question 8.

Figure 8-21 Question 4.

System's energies:

ΔK = +50 J

ΔUg = +20 J

ΔEth = ?

W = +100 J

(b)(a) (c)

Cylinder

Earth

Rope

Rod

System

Work W

C D

Figure 8-22 Question 5.

Figure 8-23 Question 6.

(

)

0.5 m

1.5 m

3.0 m

2.5 m

3.5 m

(2)

(3)

(4)

Cylinder

Rod

Block

Rope

Figure 8-24 Question 7.

5In Fig. 8-22, a block slides from Ato Calong a frictionless ramp,

and then it passes through horizontal region CD, where a frictional

force acts on it. Is the block’s kinetic energy increasing, decreasing,

or constant in (a) region AB, (b) region BC, and (c) region CD?

(d) Is the block’s mechanical energy increasing, decreasing, or

constant in those regions?

6In Fig. 8-23a, you pull upward on a rope that is attached to a

cylinder on a vertical rod. Because the cylinder fits tightly on the

rod, the cylinder slides along the rod with considerable friction.

Your force does work W100 J on the cylinder–rod–Earth

system (Fig. 8-23b).An “energy statement” for the system is shown

in Fig. 8-23c: the kinetic energy Kincreases by 50 J, and the gravita-

tional potential energy Ugincreases by 20 J.The only other change

in energy within the system is for the thermal energy Eth. What is

the change Eth?

10 Figure 8-27 shows three plums

that are launched from the same level

with the same speed. One moves

straight upward, one is launched at a

small angle to the vertical, and one is

launched along a frictionless incline.

Rank the plums according to their

speed when they reach the level of

the dashed line, greatest first.

11 When a particle moves from f

to iand from jto ialong the paths

shown in Fig. 8-28, and in the indi-

cated directions, a conservative

force does the indicated amounts

of work on it. How much work is

done on the particle by when the

particle moves directly from fto j?

(1) (2) (3)

Figure 8-27 Question 10.

20 J

⫺20 J

Figure 8-28 Question 11.

7The arrangement shown in Fig. 8-24 is similar to that in

Question 6. Here you pull downward on the rope that is attached

to the cylinder, which fits tightly on the rod. Also, as the cylinder

9Figure 8-26 shows three situations involving a plane that is not

frictionless and a block sliding along the plane.The block begins with

the same speed in all three situations and slides until the kinetic fric-

tional force has stopped it. Rank the situations according to the in-

crease in thermal energy due to the sliding,greatest first.

•3 You drop a 2.00 kg book to a friend

who stands on the ground at distance

D10.0 m below. If your friend’s out-

stretched hands are at distance d1.50 m

above the ground (Fig. 8-30), (a) how

much work Wgdoes the gravitational

force do on the book as it drops to her

hands? (b) What is the change Uin the

gravitational potential energy of the

book–Earth system during the drop? If

the gravitational potential energy Uof

that system is taken to be zero at ground

level, what is U(c) when the book is re-

leased and (d) when it reaches her

hands? Now take Uto be 100 J at

ground level and again find (e) Wg,

(f) U, (g) Uat the release point, and

(h) Uat her hands.

•4 Figure 8-31 shows a ball with mass

m0.341 kg attached to the end of a thin rod

with length L0.452 m and negligible mass.

The other end of the rod is pivoted so that the

ball can move in a vertical circle. The rod is

held horizontally as shown and then given

enough of a downward push to cause the

ball to swing down and around and just reach

the vertically up position, with zero speed

there. How much work is done on the ball by

the gravitational force from the initial point

202 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

Module 8-1 Potential Energy

•1 What is the spring constant of a spring that stores 25 J of

elastic potential energy when compressed by 7.5 cm?

•2 In Fig. 8-29, a single frictionless roller-coaster car of mass

m825 kg tops the first hill with speed v017.0 m/s at height

h42.0 m. How much work does the gravitational force do on the

car from that point to (a) point A, (b) point B, and (c) point C? If the

gravitational potential energy of the car–Earth system is taken to be

zero at C, what is its value when the car is at (d) Band (e) A? (f) If

mass mwere doubled,would the change in the gravitational potential

energy of the system between points Aand Bincrease, decrease, or

remain the same?

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to (a) the lowest point, (b) the highest point, and (c) the point on

the right level with the initial point? If the gravitational potential

energy of the ball–Earth system is taken to be zero at the initial

point, what is it when the ball reaches (d) the lowest point, (e) the

highest point, and (f) the point on the right level with the initial

point? (g) Suppose the rod were pushed harder so that the ball

passed through the highest point with a nonzero speed.Would Ug

from the lowest point to the highest point then be greater than, less

than, or the same as it was when the ball stopped at the highest

point?

•5 In Fig. 8-32, a 2.00 g ice

flake is released from the edge of a

hemispherical bowl whose radius r

is 22.0 cm. The flake–bowl contact

is frictionless. (a) How much work is

done on the flake by the gravita-

tional force during the flake’s

descent to the bottom of the bowl?

(b) What is the change in the poten-

tial energy of the flake–Earth sys-

tem during that descent? (c) If that

potential energy is taken to be zero

at the bottom of the bowl, what is its

value when the flake is released? (d) If, instead, the potential en-

ergy is taken to be zero at the release point, what is its value when

the flake reaches the bottom of the bowl? (e) If the mass of the

flake were doubled, would the magnitudes of the answers to (a)

through (d) increase, decrease, or remain the same?

••6 In Fig. 8-33, a small block of

mass m0.032 kg can slide along

the frictionless loop-the-loop, with

loop radius R12 cm. The block is

released from rest at point P,at

height h5.0Rabove the bottom

of the loop. How much work does

the gravitational force do on the

block as the block travels from point

Pto (a) point Qand (b) the top of

the loop? If the gravitational poten-

tial energy of the block–Earth sys-

tem is taken to be zero at the bot-

tom of the loop, what is that potential energy when the block is (c)

at point P, (d) at point Q, and (e) at the top of the loop? (f) If, in-

stead of merely being released, the block is given some initial

speed downward along the track, do the answers to (a) through (e)

increase, decrease, or remain the same?

••7 Figure 8-34 shows a thin rod, of length L2.00 m and neg-

ligible mass, that can pivot about one end to rotate in a vertical

circle. A ball of mass m5.00 kg is attached to the other end.

The rod is pulled aside to angle u030.0and released with

initial velocity . As the ball descends to its lowest point,

(a) how much work does the gravitational force do on it and

(b) what is the change in the gravitational potential energy of

00

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Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

•–••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday

Problems

First

hill

h/2

flake

Figure 8-32 Problems 5

and 11.

Figure 8-29 Problems 2 and 9.

Figure 8-31

Problems 4

and 14.

Figure 8-30

Problems 3 and 10.

Figure 8-33 Problems 6

and 17.

••16 A 700 g block is released from rest at height h0above a ver-

tical spring with spring constant k400 N/m and negligible mass.

The block sticks to the spring and momentarily stops after com-

pressing the spring 19.0 cm. How much work is done (a) by the

block on the spring and (b) by the spring on the block? (c) What is

the value of h0? (d) If the block were released from height 2.00h0

above the spring, what would be the maximum compression of the

spring?

••17 In Problem 6, what are the magnitudes of (a) the horizontal

component and (b) the vertical component of the net force acting

on the block at point Q? (c) At what height hshould the block be

released from rest so that it is on the verge of losing contact with

the track at the top of the loop? (On the verge of losing contact

means that the normal force on the block from the track has just

then become zero.) (d) Graph the magnitude of the normal force on

the block at the top of the loop versus initial height h, for the range

h0 to h6R.

••18 (a) In Problem 7, what is the speed of the ball at the lowest

point? (b) Does the speed increase, decrease, or remain the same if

the mass is increased?

••19 Figure 8-36 shows an 8.00 kg stone

at rest on a spring.The spring is compressed

10.0 cm by the stone. (a) What is the spring

constant? (b) The stone is pushed down an

additional 30.0 cm and released.What is the

elastic potential energy of the compressed

spring just before that release? (c) What is

the change in the gravitational potential en-

ergy of the stone–Earth system when the

stone moves from the release point to its maximum height? (d) What

is that maximum height,measured from the release point?

••20 A pendulum consists of a 2.0 kg stone swinging on a

4.0 m string of negligible mass. The stone has a speed of 8.0 m/s

when it passes its lowest point. (a) What is the speed when the

string is at 60to the vertical? (b) What is the greatest angle with

the vertical that the string will reach during the stone’s motion?

to be zero at the stone’s lowest point, what is the total mechanical

energy of the system?

••21 Figure 8-34 shows a pendulum of length L1.25 m. Its bob

(which effectively has all the mass) has speed v0when the cord makes

an angle u040.0with the vertical. (a) What is the speed of the bob

when it is in its lowest position if v08.00 m/s? What is the least

value that v0can have if the pendulum is to swing down and then up

(b) to a horizontal position, and (c) to a vertical position with the

cord remaining straight? (d) Do the answers to (b) and (c) increase,

decrease,or remain the same if u0is increased by a few degrees?

203

PROBLEMS

the ball–Earth system? (c) If the gravita-

tional potential energy is taken to be zero

at the lowest point, what is its value just as

the ball is released? (d) Do the magnitudes

of the answers to (a) through (c) increase,

decrease, or remain the same if angle u0is

increased?

••8 A 1.50 kg snowball is fired from a cliff

12.5 m high.The snowball’s initial velocity is

14.0 m/s, directed 41.0above the horizontal.

(a) How much work is done on the snowball

by the gravitational force during its flight to

the flat ground below the cliff? (b) What is

the change in the gravitational potential en-

ergy of the snowball–Earth system during

the flight? (c) If that gravitational potential

energy is taken to be zero at the height of the cliff, what is its value

when the snowball reaches the ground?

Module 8-2 Conservation of Mechanical Energy

•9 In Problem 2, what is the speed of the car at (a) point A,

(b) point B, and (c) point C? (d) How high will the car go on the

last hill, which is too high for it to cross? (e) If we substitute a sec-

ond car with twice the mass, what then are the answers to (a)

through (d)?

•10 (a) In Problem 3, what is the speed of the book when it

reaches the hands? (b) If we substituted a second book with twice

the mass, what would its speed be? (c) If, instead, the book were

thrown down, would the answer to (a) increase, decrease, or

remain the same?

•11 (a) In Problem 5, what is the speed of the flake

when it reaches the bottom of the bowl? (b) If we substituted a sec-

ond flake with twice the mass, what would its speed be? (c) If,

instead, we gave the flake an initial downward speed along the

bowl, would the answer to (a) increase, decrease, or remain the

same?

•12 (a) In Problem 8, using energy techniques rather than the

techniques of Chapter 4, find the speed of the snowball as it

reaches the ground below the cliff. What is that speed (b) if the

launch angle is changed to 41.0below the horizontal and (c) if the

mass is changed to 2.50 kg?

•13 A 5.0 g marble is fired vertically upward using a spring

gun. The spring must be compressed 8.0 cm if the marble is to just

reach a target 20 m above the marble’s position on the compressed

spring.(a) What is the change Ugin the gravitational potential en-

ergy of the marble–Earth system during the 20 m ascent?

(b) What is the change Usin the elastic potential energy of the

spring during its launch of the marble? (c) What is the spring con-

stant of the spring?

•14 (a) In Problem 4, what initial speed must be given the ball so

that it reaches the vertically upward position with zero speed? What

then is its speed at (b) the lowest point and (c) the point on the right

at which the ball is level with the initial point? (d) If the ball’s mass

were doubled, would the answers to (a) through (c) increase, de-

crease, or remain the same?

•15 In Fig. 8-35, a runaway truck with failed brakes is mov-

ing downgrade at 130 km/h just before the driver steers the truck

up a frictionless emergency escape ramp with an inclination of

u15.The truck’s mass is 1.2 104kg. (a) What minimum length

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Lmust the ramp have if the truck is to stop (momentarily) along

it? (Assume the truck is a particle, and justify that assumption.)

Does the minimum length Lincrease, decrease, or remain the same

if (b) the truck’s mass is decreased and (c) its speed is decreased?

Figure 8-34

Problems 7, 18,

and 21.

Figure 8-35 Problem 15.

Figure 8-36

Problem 19.

Uof the ball–Earth system between t0 and

t6.0 s (still free fall)?

••26 A conservative force ,

where xis in meters, acts on a particle moving

along an xaxis.The potential energy Uassociated

with this force is assigned a value of 27 J at x0.

(a) Write an expression for Uas a function of x,

with Uin joules and xin meters. (b)

What is the maximum positive poten-

tial energy? At what (c) negative

value and (d) positive value of xis the

potential energy equal to zero?

••27 Tarzan, who weighs 688 N,

swings from a cliff at the end of a vine

18 m long (Fig. 8-40). From the top of

the cliff to the bottom of the swing,he

descends by 3.2 m.The vine will break

if the force on it exceeds 950 N.

(a) Does the vine break? (b) If no,

what is the greatest force on it during

the swing? If yes, at what angle with

the vertical does it break?

:(6.0x12)i

ˆ N



is L120 cm long, has a ball

attached to one end, and is

fixed at its other end. The dis-

tance dfrom the fixed end to a

fixed peg at point Pis 75.0 cm.

When the initially stationary

ball is released with the string

horizontal as shown, it will

swing along the dashed arc.

What is its speed when it

reaches (a) its lowest point

••31 A block with mass m2.00 kg is placed against a spring

and (b) its highest point after

the string catches on the peg?

••24 A block of mass m2.0 kg is dropped

from height h40 cm onto a spring of spring

constant k1960 N/m (Fig. 8-39). Find the max-

imum distance the spring is compressed.

••25 At t0 a 1.0 kg ball is thrown from a tall

tower with . What isv

:(18 m/s)i

ˆ(24 m/s)j

••23 The string in Fig. 8-38

ILW

204 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

••22 A 60 kg skier starts from rest at height H20 m above

the end of a ski-jump ramp (Fig. 8-37) and leaves the ramp at angle

u28. Neglect the effects of air resistance and assume the ramp

is frictionless. (a) What is the maximum height hof his jump above

the end of the ramp? (b) If he increased his weight by putting on a

backpack, would hthen be greater, less,or the same?

Figure 8-38 Problems 23 and 70.

End of

ramp

Figure 8-37 Problem 22.

Figure 8-39

Problem 24.

Figure 8-40 Problem 27.

Figure 8-43 Problem 30.

••28 Figure 8-41aapplies to the spring

in a cork gun (Fig. 8-41b); it shows the

spring force as a function of the stretch or

compression of the spring. The spring is

compressed by 5.5 cm and used to propel

a 3.8 g cork from the gun. (a) What is the

speed of the cork if it is released as the

spring passes through its relaxed posi-

tion? (b) Suppose, instead, that the cork

sticks to the spring and stretches it 1.5 cm

before separation occurs.What now is the

speed of the cork at the time of release?

of mass m12 kg is released from rest

on a frictionless incline of angle 30.

Below the block is a spring that can be

compressed 2.0 cm by a force of 270 N.

The block momentarily stops when

it compresses the spring by 5.5 cm.

(a) How far does the block move

down the incline from its rest posi-

tion to this stopping point? (b) What

is the speed of the block just as it

touches the spring?

••30 A 2.0 kg breadbox on a fric-

tionless incline of angle u40is

connected, by a cord that runs over a

pulley, to a light spring of spring con-

stant k120 N/m, as shown in





Figure 8-42 Problems 29

and 35.

••29 In Fig. 8-42, a block

WWWSSM

x (cm)

2 4 –2–4

0.4

0.2

(a)

(b)

Compressed

spring

Cork

–0.2

–0.4

Force (N)

Figure 8-41 Problem 28.

Fig. 8-43. The box is released from rest when the spring is

unstretched.Assume that the pulley is massless and frictionless. (a)

What is the speed of the box when it has moved 10 cm down the in-

cline? (b) How far down the incline from its point of release does

the box slide before momentarily stopping, and what are the (c)

magnitude and (d) direction (up or down the incline) of the box’s

acceleration at the instant the box momentarily stops?

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on a frictionless incline with angle 30.0(Fig. 8-44). (The block is

not attached to the spring.) The spring, with spring constant k19.6

N/cm, is compressed 20.0 cm and then released. (a) What is the elastic

potential energy of the compressed

spring? (b) What is the change in the

gravitational potential energy of the

block–Earth system as the block

moves from the release point to its

highest point on the incline? (c)

How far along the incline is the

highest point from the release

point?







Figure 8-44 Problem 31.

a distance ddown a frictionless incline at angle u30.0where it

runs into a spring of spring constant 431 N/m.When the block mo-

mentarily stops, it has compressed the spring by 21.0 cm.What are

(a) distance dand (b) the distance between the point of the first

block–spring contact and the point where the block’s speed is

greatest?

•••36 Two children are

playing a game in which

they try to hit a small box

on the floor with a marble

fired from a spring-loaded

gun that is mounted on a

table.The target box is hori-

zontal distance D2.20 m

from the edge of the table;

see Fig. 8-48. Bobby compresses the spring 1.10 cm, but the center

of the marble falls 27.0 cm short of the center of the box. How far

should Rhoda compress the spring to score a direct hit? Assume

that neither the spring nor the ball encounters friction in the gun.

•••37 A uniform cord of length 25 cm and mass 15 g is initially

stuck to a ceiling. Later, it hangs vertically from the ceiling with only

one end still stuck. What is the change in the gravitational potential

energy of the cord with this change in orientation? (Hint: Consider a

differential slice of the cord and then use integral calculus.)

Module 8-3 Reading a Potential Energy Curve

••38 Figure 8-49 shows a plot of potential energy Uversus posi-

tion xof a 0.200 kg particle that can travel only along an xaxis

under the influence of a conservative force. The graph has these

••39 Figure 8-50 shows a

plot of potential energy Uver-

sus position xof a 0.90 kg parti-

cle that can travel only along an

xaxis. (Nonconservative forces

are not involved.) Three values

are

and The particle is

released at x4.5 m with an

initial speed of 7.0 m/s, headed

in the negative xdirection.

(a) If the particle can reach x1.0 m, what is its speed there, and if

it cannot, what is its turning point? What are the (b) magnitude

and (c) direction of the force on the particle as it begins to move to

the left of x4.0 m? Suppose, instead, the particle is headed in the

positive xdirection when it is released at x4.5 m at speed 7.0 m/s.

(d) If the particle can reach x7.0 m, what is its speed there, and if

it cannot, what is its turning point? What are the (e) magnitude and

(f) direction of the force on the particle as it begins to move to the

right of x5.0 m?

••40 The potential energy of a diatomic molecule (a two-atom

system like H2or O2) is given by

where ris the separation of the two atoms of the molecule and A

and Bare positive constants. This potential energy is associated

with the force that binds the two atoms together. (a) Find the equilib-

rium separation —that is, the distance between the atoms at which the

force on each atom is zero. Is the force repulsive (the atoms are

pushed apart) or attractive (they are pulled together) if their separa-

tion is (b) smaller and (c) larger than the equilibrium separation?

•••41 A single conservative force F(x) acts on a 1.0 kg particle

that moves along an xaxis. The potential energy U(x) associated

with F(x) is given by

U(x)4x ex/4 J,

where xis in meters.At x5.0 m the particle has a kinetic energy

of 2.0 J. (a) What is the mechanical energy of the system? (b) Make

UA

r12 B

r6,



C45.0 J.

B35.0 J,U

A15.0 J,

k170 N/m is at the top of a fric-

tionless incline of angle 37.0.

The lower end of the incline is dis-

tance D1.00 m from the end of

the spring, which is at its relaxed

length. A 2.00 kg canister is pushed

against the spring until the spring is

compressed 0.200 m and released

from rest. (a) What is the speed of

the canister at the instant the spring

returns to its relaxed length (which is when the canister loses contact

with the spring)? (b) What is the speed of the canister when it

reaches the lower end of the incline?

•••34 A boy is initially seated

on the top of a hemispherical ice

mound of radius R13.8 m. He

begins to slide down the ice, with a

negligible initial speed (Fig. 8-47).

Approximate the ice as being fric-

tionless. At what height does the

boy lose contact with the ice?

•••35 In Fig. 8-42, a block of mass m3.20 kg slides from rest







••32 In Fig. 8-45, a chain is held

on a frictionless table with one-

fourth of its length hanging over

the edge. If the chain has length

L28 cm and mass m0.012 kg,

how much work is required to pull

the hanging part back onto the

table?

•••33 In Fig. 8-46, a spring with

205

PROBLEMS

values: , and .The particle is

released at the point where Uforms a “potential hill” of “height”

, with kinetic energy 4.00 J. What is the speed of the

particle at (a) m and (b) m? What is the position

of the turning point on (c) the right side and (d) the left side?

x6.5x3.5

B12.00 J

D24.00 JU

A9.00 J, U

C20.00 J

Figure 8-46 Problem 33.

Figure 8-48 Problem 36.

Figure 8-47 Problem 34.

Figure 8-45 Problem 32.

246

x (m)

U (J)

Figure 8-50 Problem 39.

U (J)

1 2 3 4 5 6 7 8 9 0

x (m)

Figure 8-49 Problem 38.

•50 A 60 kg skier leaves the end of a ski-jump ramp with a ve-

locity of 24 m/s directed 25above the horizontal. Suppose that as a

result of air drag the skier returns to the ground with a speed of 22

m/s, landing 14 m vertically below the end of the ramp. From the

launch to the return to the ground, by how much is the mechanical

energy of the skier–Earth system reduced because of air drag?

•51 During a rockslide, a 520 kg rock slides from rest down a hill-

side that is 500 m long and 300 m high. The coefficient of kinetic

friction between the rock and the hill surface is 0.25. (a) If the grav-

itational potential energy Uof the rock–Earth system is zero at

the bottom of the hill, what is the value of Ujust before the slide?

(b) How much energy is transferred to thermal energy during the

slide? (c) What is the kinetic energy of the rock as it reaches the

bottom of the hill? (d) What is its speed then?

••52 A large fake cookie sliding on a horizontal surface is

attached to one end of a horizontal spring with spring constant

k400 N/m; the other end of the spring is fixed in place. The

cookie has a kinetic energy of 20.0 J as it passes through

the spring’s equilibrium position. As the cookie slides, a frictional

force of magnitude 10.0 N acts on it. (a) How far will the cookie

slide from the equilibrium position before coming momentarily to

rest? (b) What will be the kinetic energy of the cookie as it slides

back through the equilibrium position?

••53 In Fig. 8-52, a 3.5 kg

block is accelerated from rest

by a compressed spring of

spring constant 640 N/m. The

block leaves the spring at the

spring’s relaxed length and

then travels over a horizontal

floor with a coefficient of ki-

netic friction mk0.25. The frictional force stops the block in dis-

tance D7.8 m. What are (a) the increase in the thermal energy

of the block–floor system, (b) the maximum kinetic energy of the

block, and (c) the original compression distance of the spring?

••54 A child whose weight is 267 N slides down a 6.1 m play-

ground slide that makes an angle of 20with the horizontal.The co-

efficient of kinetic friction between slide and child is 0.10. (a) How

much energy is transferred to thermal energy? (b) If she starts at

the top with a speed of 0.457 m/s, what is her speed at the bottom?

••55 In Fig. 8-53, a block of mass m2.5 kg slides head on

into a spring of spring constant

k320 N/m. When the block

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stops, it has compressed the

spring by 7.5 cm. The coefficient

of kinetic friction between block

and floor is 0.25. While the block

is in contact with the spring and

being brought to rest, what are (a)

the work done by the spring force and (b) the increase in thermal

energy of the block–floor system? (c) What is the block’s speed

just as it reaches the spring?

••56 You push a 2.0 kg block against a horizontal spring, com-

pressing the spring by 15 cm. Then you release the block, and the

206 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

a plot of U(x) as a function of xfor 0 x10 m, and on the same

graph draw the line that represents the mechanical energy of the

system. Use part (b) to determine (c) the least value of xthe parti-

cle can reach and (d) the greatest value of xthe particle can reach.

Use part (b) to determine (e) the maximum kinetic energy of the

particle and (f) the value of xat which it occurs. (g) Determine an

expression in newtons and meters for F(x) as a function of x. (h)

For what (finite) value of xdoes F(x)0?

Module 8-4 Work Done on a System by an External Force

•42 A worker pushed a 27 kg block 9.2 m along a level floor at con-

stant speed with a force directed 32below the horizontal. If the coef-

ficient of kinetic friction between block and floor was 0.20, what were

(a) the work done by the worker’s force and (b) the increase in ther-

mal energy of the block–floor system?

•43 A collie drags its bed box across a floor by applying a hori-

zontal force of 8.0 N. The kinetic frictional force acting on the box

has magnitude 5.0 N. As the box is dragged through 0.70 m along

the way, what are (a) the work done by the collie’s applied force

and (b) the increase in thermal energy of the bed and floor?

••44 A horizontal force of magnitude 35.0 N pushes a block of

mass 4.00 kg across a floor where the coefficient of kinetic friction is

0.600. (a) How much work is done by that applied force on the

block–floor system when the block slides through a displacement of

3.00 m across the floor? (b) During that displacement, the thermal

energy of the block increases by 40.0 J. What is the increase in ther-

mal energy of the floor? (c) What is the increase in the kinetic energy

of the block?

••45 A rope is used to pull a 3.57 kg block at constant speed

4.06 m along a horizontal floor. The force on the block from the

rope is 7.68 N and directed 15.0above the horizontal.What are (a)

the work done by the rope’s force, (b) the increase in thermal en-

ergy of the block–floor system, and (c) the coefficient of kinetic

friction between the block and floor?

Module 8-5 Conservation of Energy

•46 An outfielder throws a baseball with an initial speed of

81.8 mi/h. Just before an infielder catches the ball at the same

level, the ball’s speed is 110 ft/s. In foot-pounds, by how much is the

mechanical energy of the ball–Earth system reduced because of

air drag? (The weight of a baseball is 9.0 oz.)

•47 A 75 g Frisbee is thrown from a point 1.1 m above the

ground with a speed of 12 m/s. When it has reached a height of

2.1 m, its speed is 10.5 m/s. What was the reduction in Emec of the

Frisbee–Earth system because of air drag?

•48 In Fig. 8-51, a block slides

down an incline. As it moves from

point Ato point B, which are 5.0 m

apart, force acts on the block, with

magnitude 2.0 N and directed down

the incline.The magnitude of the fric-

tional force acting on the block is

10 N. If the kinetic energy of the

block increases by 35 J between Aand B, how much work is done on

the block by the gravitational force as the block moves from Ato B?

•49 A 25 kg bear slides, from rest, 12 m down a

lodgepole pine tree, moving with a speed of 5.6 m/s just before

hitting the ground. (a) What change occurs in the gravitational

ILWSSM

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potential energy of the bear–Earth system during the slide?

(b) What is the kinetic energy of the bear just before hitting the

ground? (c) What is the average frictional force that acts on the

sliding bear?

Figure 8-53 Problem 55.

No friction D

(k)

Figure 8-52 Problem 53.

Figure 8-51 Problems 48

and 71.

•••65 A particle can slide along a

track with elevated ends and a flat

central part, as shown in Fig. 8-58.

The flat part has length L40 cm.

The curved portions of the track are

frictionless, but for the flat part the

coefficient of kinetic friction is mk

0.20.The particle is released from rest at point A, which is at height

hL/2. How far from the left edge of the flat part does the particle

finally stop?

Additional Problems

66 A 3.2 kg sloth hangs 3.0 m above the ground. (a) What is the

gravitational potential energy of the sloth–Earth system if we take

the reference point y0 to be at the ground? If the sloth drops

to the ground and air drag on it is assumed to be negligible, what

are the (b) kinetic energy and (c) speed of the sloth just before it

reaches the ground?

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207

PROBLEMS

spring sends it sliding across a tabletop. It stops 75 cm from where

you released it. The spring constant is 200 N/m. What is the

block–table coefficient of kinetic friction?

••57 In Fig. 8-54, a block slides along a track from one level to a

higher level after passing through an intermediate valley. The track

is frictionless until the block reaches the higher level. There a fric-

tional force stops the block in a distance d. The block’s initial speed

v0is 6.0 m/s,the height difference his 1.1 m, and mkis 0.60.Find d.

•••63 The cable of the 1800 kg elevator

cab in Fig. 8-56 snaps when the cab is at

rest at the first floor, where the cab bottom

is a distance d3.7 m above a spring of

spring constant k0.15 MN/m. A safety

device clamps the cab against guide rails so

that a constant frictional force of 4.4 kN

opposes the cab’s motion. (a) Find the

speed of the cab just before it hits the

spring. (b) Find the maximum distance x

that the spring is compressed (the fric-

tional force still acts during this compres-

sion). (c) Find the distance that the cab

will bounce back up the shaft. (d) Using

conservation of energy, find the approxi-

mate total distance that the cab will move before coming to rest.

(Assume that the frictional force on the cab is negligible when the

cab is stationary.)

•••64 In Fig. 8-57, a block is released from rest at height d40

cm and slides down a frictionless ramp and onto a first plateau,

which has length dand where the coefficient of kinetic friction is

0.50. If the block is still moving, it then slides down a second fric-

tionless ramp through height d/2 and onto a lower plateau, which

has length d/2 and where the coefficient of kinetic friction is

again 0.50. If the block is still moving, it then slides up a friction-

less ramp until it (momentarily) stops. Where does the block

stop? If its final stop is on a plateau, state which one and give the

distance Lfrom the left edge of that plateau. If the block reaches

the ramp, give the height Habove the lower plateau where it

momentarily stops.

= 0

Figure 8-54 Problem 57.

••58 A cookie jar is moving up a 40incline. At a point 55 cm

from the bottom of the incline (measured along the incline), the jar

has a speed of 1.4 m/s. The coefficient of kinetic friction between

jar and incline is 0.15. (a) How much farther up the incline will the

jar move? (b) How fast will it be going when it has slid back to the

bottom of the incline? (c) Do the answers to (a) and (b) increase,

decrease, or remain the same if we decrease the coefficient of ki-

netic friction (but do not change the given speed or location)?

••59 A stone with a weight of 5.29 N is launched vertically from

ground level with an initial speed of 20.0 m/s, and the air drag on it

is 0.265 N throughout the flight. What are (a) the maximum height

reached by the stone and (b) its speed just before it hits the ground?

••60 A 4.0 kg bundle starts up a 30incline with 128 J of kinetic

energy. How far will it slide up the incline if the coefficient of ki-

netic friction between bundle and incline is 0.30?

••61 When a click beetle is upside down on its back, it jumps

upward by suddenly arching its back, transferring energy stored in a

muscle to mechanical energy.This launching mechanism produces an

audible click, giving the beetle its name.Videotape of a certain click-

beetle jump shows that a beetle of mass m4.0 106kg moved di-

rectly upward by 0.77 mm during the launch and then to a maximum

height of h0.30 m. During the launch, what are the average mag-

nitudes of (a) the external force on the beetle’s back from the floor

and (b) the acceleration of the beetle in terms of g?

•••62 In Fig. 8-55, a block slides along a path that is without fric-

tion until the block reaches the section of length L0.75 m, which

begins at height h2.0 m on a ramp of angle u30. In that sec-

tion, the coefficient of kinetic friction is 0.40. The block passes

through point Awith a speed of 8.0 m/s. If the block can reach point

B(where the friction ends), what is its speed there, and if it cannot,

what is its greatest height above A?



Figure 8-55 Problem 62.

d/2

Figure 8-57 Problem 64.

Figure 8-58 Problem 65.

Figure 8-56

Problem 63.

between snow and skis would make him stop just at the top of

the lower peak?

73 The temperature of a plastic cube is monitored while the

cube is pushed 3.0 m across a floor at constant speed by a horizon-

tal force of 15 N. The thermal energy of the cube increases by 20 J.

What is the increase in the thermal energy of the floor along which

the cube slides?

74 A skier weighing 600 N goes over a frictionless circular hill

of radius R20 m (Fig. 8-62). Assume that the effects of air re-

sistance on the skier are negligible. As she comes up the hill, her

speed is 8.0 m/s at point B, at angle u20. (a) What is her

speed at the hilltop (point A) if she coasts without using her

poles? (b) What minimum speed can she have at Band still coast

to the hilltop? (c) Do the answers to these two questions in-

crease, decrease, or remain the same if the skier weighs 700 N

instead of 600 N?



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208 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

67 A spring (k200 N/m) is

fixed at the top of a frictionless plane

inclined at angle 40(Fig. 8-59). A

1.0 kg block is projected up the plane,

from an initial position that is distance

d0.60 m from the end of the relaxed

spring, with an initial kinetic energy of

16 J. (a) What is the kinetic energy of

the block at the instant it has com-

pressed the spring 0.20 m? (b) With

what kinetic energy must the block be

projected up the plane if it is to stop momentarily when it has

compressed the spring by 0.40 m?

68 From the edge of a cliff, a 0.55 kg projectile is launched

with an initial kinetic energy of 1550 J.The projectile’s maximum

upward displacement from the launch point is 140 m. What

are the (a) horizontal and (b) vertical components of its launch

velocity? (c) At the instant the vertical component of its velocity

is 65 m/s, what is its vertical displacement from the launch

point?

69 In Fig. 8-60, the pulley has

negligible mass, and both it and the

inclined plane are frictionless. Block

Ahas a mass of 1.0 kg, block Bhas a

mass of 2.0 kg, and angle uis 30.If

the blocks are released from rest

with the connecting cord taut, what

is their total kinetic energy when

block Bhas fallen 25 cm?

70 In Fig. 8-38, the string is L120 cm long, has a ball

attached to one end, and is fixed at its other end. A fixed peg is at

point P. Released from rest, the ball swings down until the string

catches on the peg; then the ball swings up, around the peg. If the

ball is to swing completely around the peg, what value must dis-

tance dexceed? (Hint: The ball must still be moving at the top of

its swing.Do you see why?)

71 In Fig. 8-51, a block is sent sliding down a frictionless

ramp. Its speeds at points Aand Bare 2.00 m/s and 2.60 m/s, re-

spectively. Next, it is again sent sliding down the ramp, but this

time its speed at point Ais 4.00 m/s. What then is its speed at

point B?

72 Two snowy peaks are at heights H850 m and h750 m

above the valley between them. A ski run extends between

the peaks, with a total length of 3.2 km and an average slope of

30(Fig. 8-61). (a) A skier starts from rest at the top of

the higher peak. At what speed will he arrive at the top of

the lower peak if he coasts without using ski poles? Ignore fric-

tion. (b) Approximately what coefficient of kinetic friction







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





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Figure 8-59 Problem 67.

Outward Inward

(a) 3.0 3.0

(b) 5.0 5.0

(d) 3.0x23.0x2

Figure 8-60 Problem 69.

θ θ

Figure 8-61 Problem 72.

Figure 8-62 Problem 74.

Find the net work done on the particle by the external force for the

round trip for each of the four situations.(e) For which, if any, is the

external force conservative?

77 A conservative force F(x) acts on a 2.0 kg particle that

moves along an xaxis. The potential energy U(x) associated with

F(x) is graphed in Fig. 8-63. When the particle is at x2.0 m, its

SSM

75 To form a pendulum, a 0.092 kg ball is attached to one end

of a rod of length 0.62 m and negligible mass, and the other end of

the rod is mounted on a pivot. The rod is rotated until it is straight

up, and then it is released from rest so that it swings down around

the pivot. When the ball reaches its lowest point, what are (a) its

speed and (b) the tension in the rod? Next, the rod is rotated until it

is horizontal, and then it is again released from rest. (c) At what an-

gle from the vertical does the tension in the rod equal the weight of

the ball? (d) If the mass of the ball is increased, does the answer to

76 We move a particle along an xaxis, first outward from x1.0 m

to x4.0 m and then back to x1.0 m, while an external force

acts on it. That force is directed along the xaxis, and its xcompo-

nent can have different values for the outward trip and for the re-

turn trip. Here are the values (in newtons) for four situations, where

xis in meters:





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209

PROBLEMS

velocity is 1.5 m/s. What are the (a) magnitude and (b) direction

of F(x) at this position? Between what positions on the (c) left and

(d) right does the particle move? (e) What is the particle’s speed at

x7.0 m?

81 A particle can move along only an xaxis, where conservative

forces act on it (Fig. 8-66 and the following table). The particle is

released at x5.00 m with a kinetic energy of K14.0 J and a

potential energy of U0. If its motion is in the negative direction

of the xaxis, what are its (a) Kand (b) Uat x2.00 m and its

the xaxis, what are its (e) Kand (f) Uat x11.0 m, its (g) Kand

(h) Uat x12.0 m, and its (i) Kand ( j) Uat x13.0 m? (k) Plot

U(x) versus xfor the range x0 to x13.0 m.







–5

–10

–15

–20

U(x) ( J)

x(m)

10 150

Figure 8-63 Problem 77.

78 At a certain factory, 300 kg

crates are dropped vertically from

a packing machine onto a conveyor

belt moving at 1.20 m/s (Fig. 8-64).

(A motor maintains the belt’s con-

stant speed.) The coefficient of ki-

netic friction between the belt and

each crate is 0.400. After a short

time, slipping between the belt and

the crate ceases, and the crate then moves along with the belt. For

the period of time during which the crate is being brought to rest

relative to the belt, calculate, for a coordinate system at rest in

the factory, (a) the kinetic energy supplied to the crate, (b) the

magnitude of the kinetic frictional force acting on the crate, and

(a) and (c) differ.

79 A 1500 kg car begins sliding down a 5.0inclined road

with a speed of 30 km/h. The engine is turned off, and the only

forces acting on the car are a net frictional force from the road and

the gravitational force. After the car has traveled 50 m along the

road, its speed is 40 km/h. (a) How much is the mechanical energy

of the car reduced because of the net frictional force? (b) What is

the magnitude of that net frictional force?

80 In Fig. 8-65, a 1400 kg block of granite is pulled up an incline

at a constant speed of 1.34 m/s by a cable and winch.The indicated

distances are d140 m and d230 m. The coefficient of kinetic

friction between the block and the incline is 0.40. What is the

power due to the force applied to the block by the cable?



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Figure 8-65 Problem 80.

FRAGILE

Figure 8-64 Problem 78.

0 2 4 6 8 10 12

F1F2F3F4

x (m)

Figure 8-66 Problems 81 and 82.

Range Force

0 to 2.00 m

2.00 m to 3.00 m

3.00 m to 8.00 m F0

8.00 m to 11.0 m

11.0 m to 12.0 m

12.0 m to 15.0 m F0

4(1.00 N)i

3(4.00 N)i



2(5.00 N)i

1(3.00 N)i

Next, the particle is released from rest at x0. What are (l) its

kinetic energy at x5.0 m and (m) the maximum positive position

xmax it reaches? (n) What does the particle do after it reaches xmax?



82 For the arrangement of forces in Problem 81, a 2.00 kg parti-

cle is released at x5.00 m with an initial velocity of 3.45 m/s in

the negative direction of the xaxis. (a) If the particle can reach

x0 m, what is its speed there, and if it cannot, what is its turning

point? Suppose, instead, the particle is headed in the positive xdi-

rection when it is released at x5.00 m at speed 3.45 m/s. (b) If

the particle can reach x13.0 m, what is its speed there, and if it

cannot, what is its turning point?

83 A 15 kg block is accelerated at 2.0 m/s2along a horizon-

tal frictionless surface, with the speed increasing from 10 m/s to

30 m/s. What are (a) the change in the block’s mechanical energy

and (b) the average rate at which energy is transferred to the

block? What is the instantaneous rate of that transfer when the

block’s speed is (c) 10 m/s and (d) 30 m/s?

84 A certain spring is found not to conform to Hooke’s law. The

force (in newtons) it exerts when stretched a distance x(in meters)

is found to have magnitude 52.8x38.4x2in the direction oppos-

ing the stretch. (a) Compute the work required to stretch the

spring from x0.500 m to x1.00 m. (b) With one end of the

spring fixed, a particle of mass 2.17 kg is attached to the other end

of the spring when it is stretched by an amount x1.00 m. If the

particle is then released from rest, what is its speed at the instant

the stretch in the spring is x0.500 m? (c) Is the force exerted by

the spring conservative or nonconservative? Explain.

85 Each second, 1200 m3of water passes over a waterfall

100 m high. Three-fourths of the kinetic energy gained by the water

in falling is transferred to electrical energy by a hydroelectric gener-

ator. At what rate does the generator produce electrical energy?

(The mass of 1 m3of water is 1000 kg.)

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



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

a spring of spring constant k200 N/m that has one end fixed, as

shown in Fig. 8-69. The horizontal

surface and the pulley are friction-

less, and the pulley has negligible

mass. The blocks are released from

rest with the spring relaxed.

(a) What is the combined kinetic

energy of the two blocks when the

hanging block has fallen 0.090 m?

(b) What is the kinetic energy of

the hanging block when it has

fallen that 0.090 m? (c) What maxi-

mum distance does the hanging block fall before momentarily

stopping?

92 A volcanic ash flow is moving across horizontal ground when

it encounters a 10upslope. The front of the flow then travels 920

m up the slope before stopping. Assume that the gases entrapped

in the flow lift the flow and thus make the frictional force from the

ground negligible; assume also that the mechanical energy of

the front of the flow is conserved.What was the initial speed of the

front of the flow?

93 A playground slide is in the form of an arc of a circle that has

a radius of 12 m.The maximum height of the slide is h4.0 m, and

the ground is tangent to the circle (Fig. 8-70). A 25 kg child starts

from rest at the top of the slide and has a speed of 6.2 m/s at the

bottom. (a) What is the length of the slide? (b) What average fric-

tional force acts on the child over this distance? If, instead of the

ground, a vertical line through the top of the slide is tangent to the

circle, what are (c) the length of the slide and (d) the average fric-

tional force on the child?

210 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

86 In Fig. 8-67, a small block is sent through point Awith a

speed of 7.0 m/s. Its path is without friction until it reaches the sec-

tion of length L12 m, where the coefficient of kinetic friction is

0.70. The indicated heights are h16.0 m and h22.0 m. What

are the speeds of the block at (a) point Band (b) point C? (c) Does

the block reach point D? If so, what is its speed there; if not, how far

through the section of friction does it travel?





91 Two blocks, of masses M2.0 kg and 2M, are connected to

C D

h2L

Figure 8-67 Problem 86.

87 A massless rigid rod of

length Lhas a ball of mass m

attached to one end (Fig. 8-68). The

other end is pivoted in such a way

that the ball will move in a vertical

circle. First, assume that there is no

friction at the pivot. The system is

launched downward from the hori-

zontal position Awith initial speed

v0. The ball just barely reaches point

Dand then stops. (a) Derive an ex-

pression for v0in terms of L,m, and

g. (b) What is the tension in the rod

when the ball passes through B? (c) A little grit is placed on the

pivot to increase the friction there. Then the ball just barely

reaches Cwhen launched from Awith the same speed as before.

What is the decrease in the mechanical energy during this motion?

(d) What is the decrease in the mechanical energy by the time the

ball finally comes to rest at Bafter several oscillations?

88 A 1.50 kg water balloon is shot straight up with an initial speed

of 3.00 m/s. (a) What is the kinetic energy of the balloon just as it is

launched? (b) How much work does the gravitational force do on

the balloon during the balloon’s full ascent? (c) What is the change

in the gravitational potential energy of the balloon–Earth system

during the full ascent? (d) If the gravitational potential energy is

taken to be zero at the launch point, what is its value when the bal-

loon reaches its maximum height? (e) If, instead, the gravitational

potential energy is taken to be zero at the maximum height, what is

its value at the launch point? (f) What is the maximum height?

89 A 2.50 kg beverage can is thrown directly downward from a

height of 4.00 m, with an initial speed of 3.00 m/s. The air drag on

the can is negligible. What is the kinetic energy of the can (a) as it

reaches the ground at the end of its fall and (b) when it is halfway

to the ground? What are (c) the kinetic energy of the can and (d)

the gravitational potential energy of the can–Earth system 0.200 s

before the can reaches the ground? For the latter, take the refer-

ence point y0 to be at the ground.

90 A constant horizontal force moves a 50 kg trunk 6.0 m up a

30incline at constant speed. The coefficient of kinetic friction is

0.20. What are (a) the work done by the applied force and (b) the

increase in the thermal energy of the trunk and incline?

SSM

Figure 8-69 Problem 91.

Figure 8-70 Problem 93.

ALC

Pivot

point

Rod

Figure 8-68 Problem 87.

94 The luxury liner Queen Elizabeth 2 has a diesel-electric

power plant with a maximum power of 92 MW at a cruising speed

of 32.5 knots. What forward force is exerted on the ship at this

speed? (1 knot 1.852 km/h.)

95 A factory worker accidentally releases a 180 kg crate that was

being held at rest at the top of a ramp that is 3.7 m long and in-

clined at 39to the horizontal.The coefficient of kinetic friction be-

tween the crate and the ramp, and between the crate and the hori-

zontal factory floor, is 0.28. (a) How fast is the crate moving as it

reaches the bottom of the ramp? (b) How far will it subsequently

slide across the floor? (Assume that the crate’s kinetic energy does

not change as it moves from the ramp onto the floor.) (c) Do the

answers to (a) and (b) increase, decrease, or remain the same if we

halve the mass of the crate?

96 If a 70 kg baseball player steals home by sliding into the plate

with an initial speed of 10 m/s just as he hits the ground, (a) what

211

PROBLEMS

is the decrease in the player’s kinetic energy and (b) what is the

increase in the thermal energy of his body and the ground along

which he slides?

97 A 0.50 kg banana is thrown directly upward with an initial

speed of 4.00 m/s and reaches a maximum height of 0.80 m. What

change does air drag cause in the mechanical energy of the

banana–Earth system during the ascent?

98 A metal tool is sharpened by being held against the rim of a

wheel on a grinding machine by a force of 180 N. The frictional

forces between the rim and the tool grind off small pieces of the

tool. The wheel has a radius of 20.0 cm and rotates at 2.50 rev/s.

The coefficient of kinetic friction between the wheel and the tool is

0.320. At what rate is energy being transferred from the motor

driving the wheel to the thermal energy of the wheel and tool and

to the kinetic energy of the material thrown from the tool?

99 A swimmer moves through the water at an average speed of

0.22 m/s. The average drag force is 110 N. What average power is

required of the swimmer?

100 An automobile with passengers has weight 16 400 N and is

moving at 113 km/h when the driver brakes, sliding to a stop. The

frictional force on the wheels from the road has a magnitude of

8230 N. Find the stopping distance.

101 A 0.63 kg ball thrown directly upward with an initial speed

of 14 m/s reaches a maximum height of 8.1 m. What is the change

in the mechanical energy of the ball–Earth system during the

ascent of the ball to that maximum height?

102 The summit of Mount Everest is 8850 m above sea level.

(a) How much energy would a 90 kg climber expend against the

gravitational force on him in climbing to the summit from sea

level? (b) How many candy bars, at 1.25 MJ per bar, would supply

an energy equivalent to this? Your answer should suggest that

work done against the gravitational force is a very small part of the

energy expended in climbing a mountain.

103 A sprinter who weighs 670 N runs the first 7.0 m of a race in

1.6 s, starting from rest and accelerating uniformly. What are the

sprinter’s (a) speed and (b) kinetic energy at the end of the 1.6 s?

interval?

104 A 20 kg object is acted on by a conservative force given by

F3.0x5.0x2, with Fin newtons and xin meters. Take the

potential energy associated with the force to be zero when the

object is at x0. (a) What is the potential energy of the system

associated with the force when the object is at x2.0 m? (b) If

the object has a velocity of 4.0 m/s in the negative direction of the

xaxis when it is at x5.0 m, what is its speed when it passes

through the origin? (c) What are the answers to (a) and (b) if the

potential energy of the system is taken to be 8.0 J when the ob-

ject is at x0?

105 A machine pulls a 40 kg trunk 2.0 m up a 40ramp at con-

stant velocity, with the machine’s force on the trunk directed paral-

lel to the ramp. The coefficient of kinetic friction between the

trunk and the ramp is 0.40. What are (a) the work done on the

trunk by the machine’s force and (b) the increase in thermal en-

ergy of the trunk and the ramp?

106 The spring in the muzzle of a child’s spring gun has a spring

constant of 700 N/m.To shoot a ball from the gun, first the spring is

compressed and then the ball is placed on it. The gun’s trigger then

releases the spring, which pushes the ball through the muzzle. The

ball leaves the spring just as it leaves the outer end of the muzzle.

When the gun is inclined upward by 30to the horizontal,a 57 g ball

is shot to a maximum height of 1.83 m above the gun’s muzzle.

Assume air drag on the ball is negligible. (a) At what speed does

the spring launch the ball? (b) Assuming that friction on the ball

within the gun can be neglected, find the spring’s initial compres-

sion distance.

107 The only force acting on a particle is conservative force . If

the particle is at point A, the potential energy of the system associ-

ated with and the particle is 40 J. If the particle moves from point

Ato point B, the work done on the particle by is 25 J. What is

the potential energy of the system with the particle at B?

108 In 1981, Daniel Goodwin climbed 443 m up the exterior of

the Sears Building in Chicago using suction cups and metal clips.

(a) Approximate his mass and then compute how much energy he

had to transfer from biomechanical (internal) energy to the gravi-

tational potential energy of the Earth–Goodwin system to lift

himself to that height. (b) How much energy would he have had to

transfer if he had, instead, taken the stairs inside the building (to

the same height)?

109 A 60.0 kg circus performer slides 4.00 m down a pole to the

circus floor, starting from rest. What is the kinetic energy of the

performer as she reaches the floor if the frictional force on her

from the pole (a) is negligible (she will be hurt) and (b) has a mag-

nitude of 500 N?

110 A 5.0 kg block is projected at 5.0 m/s up a plane that is

inclined at 30with the horizontal. How far up along the

plane does the block go (a) if the plane is frictionless and (b) if the

coefficient of kinetic friction between the block and the plane is

0.40? (c) In the latter case, what is the increase in thermal energy

of block and plane during the block’s ascent? (d) If the block then

slides back down against the frictional force, what is the block’s

speed when it reaches the original projection point?

111 A 9.40 kg projectile is fired vertically upward. Air drag de-

creases the mechanical energy of the projectile–Earth system by

68.0 kJ during the projectile’s ascent. How much higher would the

projectile have gone were air drag negligible?

112 A 70.0 kg man jumping from a window lands in an elevated

fire rescue net 11.0 m below the window. He momentarily stops

when he has stretched the net by 1.50 m. Assuming that mechani-

cal energy is conserved during this process and that the net func-

tions like an ideal spring,find the elastic potential energy of the net

when it is stretched by 1.50 m.

113 A 30 g bullet moving a horizontal velocity of 500 m/s comes

to a stop 12 cm within a solid wall. (a) What is the change in the

bullet’s mechanical energy? (b) What is the magnitude of the aver-

age force from the wall stopping it?

114 A 1500 kg car starts from rest on a horizontal road and

gains a speed of 72 km/h in 30 s. (a) What is its kinetic energy at

the end of the 30 s? (b) What is the average power required of the

car during the 30 s interval? (c) What is the instantaneous power

at the end of the 30 s interval, assuming that the acceleration is

constant?

115 A 1.50 kg snowball is shot upward at an angle of 34.0to the

horizontal with an initial speed of 20.0 m/s. (a) What is its initial

kinetic energy? (b) By how much does the gravitational potential

124 The magnitude of the gravitational force between a particle

of mass m1and one of mass m2is given by

F(x)

where Gis a constant and xis the distance between the particles.

(a) What is the corresponding potential energy function U(x)?

Assume that U(x) 0 as xand that xis positive. (b) How

much work is required to increase the separation of the particles

from xx1to xx1d?

125 Approximately 5.5 106kg of water falls 50 m over

Niagara Falls each second. (a) What is the decrease in the gravi-

tational potential energy of the water–Earth system each sec-

ond? (b) If all this energy could be converted to electrical

energy (it cannot be), at what rate would electrical energy be

supplied? (The mass of 1 m3of water is 1000 kg.) (c) If the elec-

trical energy were sold at 1 cent/kW h, what would be the yearly

income?

126 To make a pendulum, a 300 g ball is attached to one end of

a string that has a length of 1.4 m and negligible mass. (The other

end of the string is fixed.) The ball is pulled to one side until the

string makes an angle of 30.0with the vertical; then (with

the string taut) the ball is released from rest. Find (a) the speed of

the ball when the string makes an angle of 20.0with the vertical

and (b) the maximum speed of the ball. (c) What is the angle be-

tween the string and the vertical when the speed of the ball is

one-third its maximum value?

127 In a circus act, a 60 kg clown is shot from a cannon with an

initial velocity of 16 m/s at some unknown angle above the hori-

zontal.A short time later the clown lands in a net that is 3.9 m ver-

tically above the clown’s initial position. Disregard air drag. What

is the kinetic energy of the clown as he lands in the net?

128 A 70 kg firefighter slides, from rest, 4.3 m down a vertical

pole. (a) If the firefighter holds onto the pole lightly, so that the

frictional force of the pole on her is negligible, what is her speed

just before reaching the ground floor? (b) If the firefighter grasps

the pole more firmly as she slides, so that the average frictional

force of the pole on her is 500 N upward, what is her speed just be-

fore reaching the ground floor?

129 The surface of the continental United States has an area of

about 8 106km2and an average elevation of about 500 m

(above sea level). The average yearly rainfall is 75 cm. The frac-

tion of this rainwater that returns to the atmosphere by evapora-

tion is ; the rest eventually flows into the ocean. If the decrease

in gravitational potential energy of the water–Earth system asso-

ciated with that flow could be fully converted to electrical en-

ergy, what would be the average power? (The mass of 1 m3of

water is 1000 kg.)

130 A spring with spring constant k200 N/m is suspended

vertically with its upper end fixed to the ceiling and its lower

end at position y0. A block of weight 20 N is attached to the

lower end, held still for a moment, and then released. What are

(a) the kinetic energy K, (b) the change (from the initial value)

in the gravitational potential energy Ug, and (c) the change in

the elastic potential energy Ueof the spring–block system when

the block is at y5.0 cm? What are (d) K,(e) Ug, and (f) Ue

when y10 cm, (g) K, (h) Ug, and (i) Uewhen y15 cm,

and (j) K, (k) Ug, and (l) Uewhen y 20 cm?



::

Gm1m2

x2,

energy of the snowball–Earth system change as the snowball

moves from the launch point to the point of maximum height? (c)

What is that maximum height?

116 A 68 kg sky diver falls at a constant terminal speed of

59 m/s. (a) At what rate is the gravitational potential energy of the

Earth–sky diver system being reduced? (b) At what rate is the sys-

tem’s mechanical energy being reduced?

117 A 20 kg block on a horizontal surface is attached to a hori-

zontal spring of spring constant k4.0 kN/m. The block is pulled

to the right so that the spring is stretched 10 cm beyond its relaxed

length, and the block is then released from rest.The frictional force

between the sliding block and the surface has a magnitude of 80 N.

(a) What is the kinetic energy of the block when it has moved

2.0 cm from its point of release? (b) What is the kinetic energy of

the block when it first slides back through the point at which the

spring is relaxed? (c) What is the maximum kinetic energy attained

by the block as it slides from its point of release to the point at

which the spring is relaxed?

118 Resistance to the motion of an automobile consists of road

friction, which is almost independent of speed, and air drag, which

is proportional to speed-squared. For a certain car with a weight of

12 000 N, the total resistant force Fis given by F300 1.8v2,

with Fin newtons and vin meters per second. Calculate the power

212 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY

(in horsepower) required to accelerate the car at 0.92 m/s2when

the speed is 80 km/h.

119 A 50 g ball is thrown from a window with an initial

SSM

velocity of 8.0 m/s at an angle of 30above the horizontal. Using

energy methods, determine (a) the kinetic energy of the ball at the

top of its flight and (b) its speed when it is 3.0 m below the window.

Does the answer to (b) depend on either (c) the mass of the ball or

(d) the initial angle?

120 A spring with a spring constant of 3200 N/m is initially

stretched until the elastic potential energy of the spring is 1.44 J.

(U0 for the relaxed spring.) What is Uif the initial stretch is

changed to (a) a stretch of 2.0 cm, (b) a compression of 2.0 cm, and

121 A locomotive with a power capability of 1.5 MW can

accelerate a train from a speed of 10 m/s to 25 m/s in 6.0 min. (a)

Calculate the mass of the train. Find (b) the speed of the train and

during the 6.0 min interval. (d) Find the distance moved by the

train during the interval.

122 A 0.42 kg shuffleboard disk is initially at rest when a

player uses a cue to increase its speed to 4.2 m/s at constant ac-

celeration. The acceleration takes place over a 2.0 m distance, at

the end of which the cue loses contact with the disk. Then the

disk slides an additional 12 m before stopping. Assume that the

shuffleboard court is level and that the force of friction on the

disk is constant.What is the increase in the thermal energy of the

disk–court system (a) for that additional 12 m and (b) for the

entire 14 m distance? (c) How much work is done on the disk by

the cue?

123 A river descends 15 m through rapids. The speed of the wa-

ter is 3.2 m/s upon entering the rapids and 13 m/s upon leaving.

What percentage of the gravitational potential energy of the

water–Earth system is transferred to kinetic energy during the de-

scent? (Hint: Consider the descent of, say, 10 kg of water.)

SSM

131 Fasten one end of a vertical spring to a ceiling, attach a cab-

bage to the other end, and then slowly lower the cabbage until the

upward force on it from the spring balances the gravitational force

on it. Show that the loss of gravitational potential energy of the

cabbage–Earth system equals twice the gain in the spring’s poten-

tial energy.

132 The maximum force you can exert on an object with one of

your back teeth is about 750 N. Suppose that as you gradually bite

on a clump of licorice, the licorice resists compression by one of

your teeth by acting like a spring for which k2.5 105N/m. Find

(a) the distance the licorice is compressed by your tooth and

(b) the work the tooth does on the licorice during the compression.

distance. (d) If there is a potential energy associated with this com-

pression, plot it versus compression distance.

In the 1990s the pelvis of a particular Triceratops dinosaur was

found to have deep bite marks. The shape of the marks suggested

that they were made by a Tyrannosaurus rex dinosaur. To test the

idea, researchers made a replica of a T. rex tooth from bronze and

aluminum and then used a hydraulic press to gradually drive the

replica into cow bone to the depth seen in the Triceratops bone. A

graph of the force required versus depth of penetration is given in

Fig. 8-71 for one trial; the required force increased with depth be-

cause, as the nearly conical tooth penetrated the bone, more of the

tooth came in contact with the bone. (e) How much work was done

by the hydraulic press—and thus presumably by the T. rex—in

such a penetration? (f) Is there a potential energy associated with

this penetration? (The large biting force and energy expenditure

attributed to the T. rex by this research suggest that the animal was

a predator and not a scavenger.)

133 Conservative force F(x)

acts on a particle that moves

along an xaxis. Figure 8-72

shows how the potential energy

U(x) associated with force F(x)

varies with the position of the

particle, (a) Plot F(x) for the

range 0 x6 m. (b) The me-

chanical energy Eof the system

is 4.0 J. Plot the kinetic energy

K(x) of the particle directly on

Fig. 8-72.

134 Figure 8-73ashows a mol-

ecule consisting of two atoms of

masses mand M(with mM)

and separation r. Figure 8-73b

shows the potential energy U(r)

of the molecule as a function of

r. Describe the motion of the

atoms (a) if the total mechanical

energy Eof the two-atom sys-

tem is greater than zero (as is

E1), and (b) if Eis less than zero

(as is E2). For E111019 J

and r0.3 nm, find (c) the po-

tential energy of the system, (d)

the total kinetic energy of the

atoms, and (e) the force (magni-

tude and direction) acting on

each atom. For what values of r

is the force (f) repulsive, (g) at-

tractive, and (h) zero?

135 Repeat Problem 83, but now with the block accelerated up a

frictionless plane inclined at 5.0to the horizontal.

136 A spring with spring constant k620 N/m is placed in a ver-

tical orientation with its lower end supported by a horizontal sur-

face. The upper end is depressed 25 cm, and a block with a weight

of 50 N is placed (unattached) on the depressed spring.The system

is then released from rest. Assume that the gravitational potential

energy Ugof the block is zero at the release point (y0) and cal-

culate the kinetic energy Kof the block for yequal to (a) 0,

(b) 0.050 m, (c) 0.10 m, (d) 0.15 m, and (e) 0.20 m.Also, (f) how far

above its point of release does the block rise?



213

PROBLEMS

2000

4000

6000

8000

1 2 3 4 5 6

Penetration depth (mm)

7 8 9 10 11 12

Force (N)

Figure 8-71 Problem 132.

U(x) ( J)

(

)

4 5 6 10

Figure 8-72 Problem 133.

U(r), E (10–

9J)

–1

–2

–3

0.2

r (nm)

(

)

0.3 0.4 0.10

(a)

Figure 8-73 Problem 134.

214

What Is Physics?

Every mechanical engineer who is hired as a courtroom expert witness to recon-

struct a traffic accident uses physics. Every dance trainer who coaches a ballerina

on how to leap uses physics. Indeed, analyzing complicated motion of any sort re-

quires simplification via an understanding of physics. In this chapter we discuss

how the complicated motion of a system of objects, such as a car or a ballerina,

can be simplified if we determine a special point of the system—the center of

mass of that system.

Here is a quick example. If you toss a ball into the air without much spin on the

ball (Fig. 9-1a), its motion is simple—it follows a parabolic path, as we discussed in

Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat

into the air (Fig.9-1b), its motion is more complicated. Because every part of the bat

moves differently, along paths of many different shapes, you cannot represent the

bat as a particle. Instead, it is a system of particles each of which follows its own path

through the air. However, the bat has one special point—the center of mass—that

does move in a simple parabolic path. The other parts of the bat move around the

center of mass. (To locate the center of mass, balance the bat on an outstretched fin-

ger;the point is above your finger, on the bat’s central axis.)

You cannot make a career of flipping baseball bats into the air, but you can

make a career of advising long-jumpers or dancers on how to leap properly into

the air while either moving their arms and legs or rotating their torso. Your

starting point would be to determine the person’s center of mass because of its

simple motion.

CHAPTER 9

Center of Mass and Linear Momentum

9-1 CENTER OF MASS

After reading this module, you should be able to . . .

9.01 Given the positions of several particles along an axis or

a plane, determine the location of their center of mass.

9.02 Locate the center of mass of an extended, symmetric

object by using the symmetry.

9.03 For a two-dimensional or three-dimensional extended ob-

ject with a uniform distribution of mass, determine the center

of mass by (a) mentally dividing the object into simple geomet-

ric figures, each of which can be replaced by a particle at its

center and (b) finding the center of mass of those particles.

●The center of mass of a system of nparticles is defined to be the point whose coordinates are given by

where Mis the total mass of the system.

com 1



i1

mir

xcom 1



i1

mixi, ycom 1



i1

miyi, zcom 1



i1

mizi,

Key Idea

Learning Objectives

215

9-1 CENTER OF MASS

The Center of Mass

We define the center of mass (com) of a system of particles (such as a person) in

order to predict the possible motion of the system.

Here we discuss how to determine where the center of mass of a system of parti-

cles is located.We start with a system of only a few particles, and then we consider

a system of a great many particles (a solid body, such as a baseball bat). Later in

the chapter, we discuss how the center of mass of a system moves when external

forces act on the system.

Systems of Particles

Two Particles. Figure 9-2ashows two particles of masses m1and m2separated by dis-

tance d.We have arbitrarily chosen the origin of an xaxis to coincide with the particle

of mass m1.We define the position of the center of mass (com) of this two-particle sys-

tem to be

(9-1)

Suppose, as an example, that m20.Then there is only one particle, of mass m1,

and the center of mass must lie at the position of that particle;Eq. 9-1 dutifully reduces

to xcom 0. If m10,there is again only one particle (of mass m2), and we have, as we

expect, xcom d. If m1m2, the center of mass should be halfway between the two

particles; Eq. 9-1 reduces to again as we expect. Finally, Eq. 9-1 tells us that

if neither m1nor m2is zero, xcom can have only values that lie between zero and d; that

is, the center of mass must lie somewhere between the two particles.

We are not required to place the origin of the coordinate system on one of

the particles. Figure 9-2bshows a more generalized situation, in which the coordi-

nate system has been shifted leftward. The position of the center of mass is now

defined

as (9-2)

Note that if we put x10, then x2becomes dand Eq. 9-2 reduces to Eq. 9-1, as

it must. Note also that in spite of the shift of the coordinate system, the center

xcom m1x1m2x2

m1m2

xcom 1

2d,

xcom m2

m1m2

Figure 9-1 (a) A ball tossed into the air

follows a parabolic path. (b) The center

of mass (black dot) of a baseball bat

flipped into the air follows a parabolic

path, but all other points of the bat

follow more complicated curved paths.

(a)

(

)

Richard Megna/Fundamental Photographs

The center of mass of a system of particles is the point that moves as though

(1) all of the system’s mass were concentrated there and (2) all external forces

were applied there.

Figure 9-2 (a) Two particles of masses m1and m2are separated by distance d. The dot

labeled com shows the position of the center of mass, calculated from Eq. 9-1. (b) The

same as (a) except that the origin is located farther from the particles. The position of

the center of mass is calculated from Eq. 9-2. The location of the center of mass with

respect to the particles is the same in both cases.

xcom

x1d

com

m1m2

(b)

xcom

com

m1m2

(a)

This is the center of mass

of the two-particle system.

Shifting the axis

does not change

the relative position

of the com.

of mass is still the same distance from each particle. The com is a property of the

physical particles, not the coordinate system we happen to use.

We can rewrite Eq. 9-2 as

(9-3)

in which Mis the total mass of the system. (Here, Mm1m2.)

Many Particles. We can extend this equation to a more general situation in

which nparticles are strung out along the xaxis. Then the total mass is Mm1

m2mn, and the location of the center of mass is

(9-4)

The subscript iis an index that takes on all integer values from 1 to n.

Three Dimensions. If the particles are distributed in three dimensions, the cen-

ter of mass must be identified by three coordinates. By extension of Eq. 9-4, they are

(9-5)

We can also define the center of mass with the language of vectors. First

recall that the position of a particle at coordinates xi,yi, and ziis given by a posi-

tion vector (it points from the origin to the particle):

(9-6)

Here the index identifies the particle, and i

ˆ,ˆ

j, and ˆ

k are unit vectors pointing,

respectively, in the positive direction of the x,y, and zaxes. Similarly, the position

of the center of mass of a system of particles is given by a position vector:

(9-7)

If you are a fan of concise notation, the three scalar equations of Eq. 9-5 can now

be replaced by a single vector equation,

(9-8)

where again Mis the total mass of the system. You can check that this equation

is correct by substituting Eqs. 9-6 and 9-7 into it, and then separating out the x,

y, and zcomponents.The scalar relations of Eq. 9-5 result.

Solid Bodies

An ordinary object, such as a baseball bat, contains so many particles (atoms)

that we can best treat it as a continuous distribution of matter. The “particles”

then become differential mass elements dm, the sums of Eq. 9-5 become inte-

grals, and the coordinates of the center of mass are defined as

(9-9)

where Mis now the mass of the object.The integrals effectively allow us to use Eq.

9-5 for a huge number of particles, an effort that otherwise would take many years.

Evaluating these integrals for most common objects (such as a television set or

a moose) would be difficult, so here we consider only uniform objects. Such objects

have uniform density, or mass per unit volume; that is, the density r(Greek letter

xcom 1

Mxdm, ycom 1

Mydm, zcom 1

Mzdm,

rcom

:1



i1

miri

rcom

:xcomˆ

iycomˆ

jzcom ˆ

:xii

ˆyij

ˆzik

ˆ.

xcom 1



i1

mixi, ycom 1



i1

miyi, zcom 1



i1

mizi.

1



i1

mixi.

xcom m1x1m2x2m3x3  mnxn



xcom m1x1m2x2

216 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

rho) is the same for any given element of an object as for the whole object. From

Eq. 1-8,we can write

(9-10)

where dV is the volume occupied by a mass element dm, and Vis the total vol-

ume of the object. Substituting dm (M/V)dV from Eq. 9-10 into Eq. 9-9 gives

(9-11)

Symmetry as a Shortcut. You can bypass one or more of these integrals if

an object has a point, a line, or a plane of symmetry. The center of mass of such

an object then lies at that point, on that line, or in that plane. For example, the

center of mass of a uniform sphere (which has a point of symmetry) is at the

center of the sphere (which is the point of symmetry). The center of mass of a

uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The

center of mass of a banana (which has a plane of symmetry that splits it into two

equal parts) lies somewhere in the plane of symmetry.

The center of mass of an object need not lie within the object. There is no

dough at the com of a doughnut, and no iron at the com of a horseshoe.

xcom 1

VxdV, ycom 1

VydV, zcom 1

VzdV.

rdm

dV M

217

9-1 CENTER OF MASS

sides (Fig. 9-3). The three particles then have the following

coordinates:

Particle Mass (kg) x(cm) y(cm)

1 1.2 0 0

2 2.5 140 0

3 3.4 70 120

The total mass Mof the system is 7.1 kg.

From Eq. 9-5, the coordinates of the center of mass are

(Answer)

and

(Answer)

In Fig. 9-3, the center of mass is located by the position vec-

tor , which has components xcom and ycom. If we had

chosen some other orientation of the coordinate system,

these coordinates would be different but the location of the

com relative to the particles would be the same.

com

58 cm.

(1.2 kg)(0) (2.5 kg)(0) (3.4 kg)(120 cm)

7.1 kg

ycom 1



i1

miyim1y1m2y2m3y3

83 cm

(1.2 kg)(0) (2.5 kg)(140 cm) (3.4 kg)(70 cm)

7.1 kg

xcom 1



i1

mixim1x1m2x2m3x3

Sample Problem 9.01 com of three particles

Three particles of masses m11.2 kg, m22.5 kg, and

m33.4 kg form an equilateral triangle of edge length

a140 cm.Where is the center of mass of this system?

KEY IDEA

We are dealing with particles instead of an extended solid

body, so we can use Eq. 9-5 to locate their center of mass.

The particles are in the plane of the equilateral triangle, so

we need only the first two equations.

Calculations: We can simplify the calculations by choosing

the xand yaxes so that one of the particles is located at the

origin and the xaxis coincides with one of the triangle’s

Figure 9-3 Three particles form an equilateral triangle of edge

length a. The center of mass is located by the position vector .r

com

100

150

com

xcom

rcom

a a

This is the position

vector rcom for the

com (it points from

the origin to the com).

Additional examples, video, and practice available at WileyPLUS

218 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Center Location

Plate of Mass of com Mass

PcomPxP?mP

ScomSxSRm

CcomCxC0mCmSmP

Assume that mass mSof disk Sis concentrated in a parti-

cle at xSR, and mass mPis concentrated in a particle

at xP(Fig. 9-4d). Next we use Eq. 9-2 to find the center of

mass xSPof the two-particle system:

(9-12)

Next note that the combination of disk Sand plate Pis

composite plate C. Thus, the position xSPof comSPmust

coincide with the position xCof comC, which is at the origin; so

xSPxC0.Substituting this into Eq. 9-12,we get

(9-13)

We can relate these masses to the face areas of Sand Pby

noting that

mass density volume

density thickness area.

Then

Because the plate is uniform, the densities and thicknesses

are equal; we are left with

Substituting this and xSRinto Eq. 9-13, we have

(Answer)xP1

3R.

pR2

p(2R)2pR21

areaS

areaP

areaS

areaCareaS

densityS

densityP

thicknessS

thicknessP

areaS

areaP

xPxS

xSPmSxSmPxP

mSmP

Sample Problem 9.02 com of plate with missing piece

This sample problem has lots of words to read, but they will

allow you to calculate a com using easy algebra instead of

challenging integral calculus. Figure 9-4ashows a uniform

metal plate Pof radius 2Rfrom which a disk of radius Rhas

been stamped out (removed) in an assembly line.The disk is

shown in Fig. 9-4b. Using the xy coordinate system shown,

locate the center of mass comPof the remaining plate.

KEY IDEAS

(1) Let us roughly locate the center of plate Pby using sym-

metry. We note that the plate is symmetric about the xaxis

(we get the portion below that axis by rotating the upper

portion about the axis). Thus, comPmust be on the xaxis.

The plate (with the disk removed) is not symmetric about

the yaxis. However, because there is somewhat more mass

on the right of the yaxis, comPmust be somewhat to the

right of that axis. Thus, the location of comPshould be

roughly as indicated in Fig.9-4a.

(2) Plate Pis an extended solid body, so in principle we

can use Eqs. 9-11 to find the actual coordinates of the center

of mass of plate P. Here we want the xy coordinates of the

center of mass because the plate is thin and uniform. If it

had any appreciable thickness, we would just say that the

center of mass is midway across the thickness. Still, using

Eqs. 9-11 would be challenging because we would need a

function for the shape of the plate with its hole, and then we

would need to integrate the function in two dimensions.

(3) Here is a much easier way: In working with centers

of mass, we can assume that the mass of a uniform object (as

we have here) is concentrated in a particle at the object’s

center of mass.Thus we can treat the object as a particle and

avoid any two-dimensional integration.

Calculations: First, put the stamped-out disk (call it disk S)

back into place (Fig. 9-4c) to form the original composite

plate (call it plate C). Because of its circular symmetry, the

center of mass comSfor disk Sis at the center of S, at x

R(as shown). Similarly, the center of mass comCfor com-

posite plate Cis at the center of C, at the origin (as shown).

We then have the following:

Additional examples, video, and practice available at WileyPLUS

Checkpoint 1

The figure shows a uniform square plate from which four identical

squares at the corners will be removed. (a) Where is the center of mass of

the plate originally? Where is it after the removal of (b) square 1; (c)

squares 1 and 2; (d) squares 1 and 3; (e) squares 1,2, and 3; (f) all four

squares? Answer in terms of quadrants, axes,or points (without calcula-

tion, of course).

1 2

4 3

219

9-1 CENTER OF MASS

The com of the composite

plate is the same as the

com of the two pieces.

Plate P

comP

comC

comS

Disk S

Composite plate

C = S + P

(a)

(b)

(c)

(d)x

comP

comC

comS

Disk particle Plate particle

Assume the plate's

mass is concentrated

as a particle at the

plate's center of mass.

Here too, assume the

mass is concentrated

as a particle at the

center of mass.

Here too.

Here are those

three particles.

Figure 9-4 (a) Plate Pis a metal plate of radius 2R, with a circular hole of radius R.The center of mass of Pis at point comP.(b) Disk S.

(c) Disk Shas been put back into place to form a composite plate C. The center of mass comSof disk Sand the center of mass comC

of plate Care shown. (d) The center of mass comSPof the combination of Sand Pcoincides with comC, which is at x0.

Newton’s Second Law for a System of Particles

Now that we know how to locate the center of mass of a system of particles, we

discuss how external forces can move a center of mass. Let us start with a simple

system of two billiard balls.

If you roll a cue ball at a second billiard ball that is at rest, you expect that the

two-ball system will continue to have some forward motion after impact. You

would be surprised, for example, if both balls came back toward you or if both

moved to the right or to the left. You already have an intuitive sense that some-

thing continues to move forward.

What continues to move forward, its steady motion completely unaf-

fected by the collision, is the center of mass of the two-ball system. If you fo-

cus on this point—which is always halfway between these bodies because

they have identical masses—you can easily convince yourself by trial at a bil-

liard table that this is so. No matter whether the collision is glancing, head-on,

or somewhere in between, the center of mass continues to move forward, as if

the collision had never occurred. Let us look into this center-of-mass motion

in more detail.

Motion of a System’s com. To do so, we replace the pair of billiard balls with

a system of nparticles of (possibly) different masses. We are interested not in the

individual motions of these particles but only in the motion of the center of mass

of the system. Although the center of mass is just a point, it moves like a particle

whose mass is equal to the total mass of the system; we can assign a position, a ve-

locity, and an acceleration to it. We state (and shall prove next) that the vector

equation that governs the motion of the center of mass of such a system of parti-

cles is

(system of particles). (9-14)

This equation is Newton’s second law for the motion of the center of mass of

a system of particles. Note that its form is the same as the form of the equation

net Ma

com

220 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES

After reading this module, you should be able to . . .

9.04 Apply Newton’s second law to a system of particles by re-

lating the net force (of the forces acting on the particles) to

the acceleration of the system’s center of mass.

9.05 Apply the constant-acceleration equations to the motion

of the individual particles in a system and to the motion of

the system’s center of mass.

9.06 Given the mass and velocity of the particles in a system,

calculate the velocity of the system’s center of mass.

9.07 Given the mass and acceleration of the particles in a

system, calculate the acceleration of the system’s center

of mass.

9.08 Given the position of a system’s center of mass as a func-

tion of time, determine the velocity of the center of mass.

9.09 Given the velocity of a system’s center of mass as a

function of time, determine the acceleration of the center

of mass.

9.10 Calculate the change in the velocity of a com by integrat-

ing the com’s acceleration function with respect to time.

9.11 Calculate a com’s displacement by integrating the

com’s velocity function with respect to time.

9.12 When the particles in a two-particle system move with-

out the system’s com moving, relate the displacements of

the particles and the velocities of the particles.

●The motion of the center of mass of any system of particles

is governed by Newton’s second law for a system of parti-

cles, which is

net Ma

com

Here is the net force of all the external forces acting on

the system, Mis the total mass of the system, and is the

acceleration of the system’s center of mass.

:com

net

Learning Objectives

Key Idea

for the motion of a single particle. However, the three quantities that

appear in Eq. 9-14 must be evaluated with some care:

1. is the net force of all external forces that act on the system. Forces on one

part of the system from another part of the system (internal forces) are not in-

cluded in Eq. 9-14.

2. Mis the total mass of the system. We assume that no mass enters or leaves the

system as it moves, so that Mremains constant. The system is said to be closed.

3. is the acceleration of the center of mass of the system. Equation 9-14 gives

no information about the acceleration of any other point of the system.

Equation 9-14 is equivalent to three equations involving the components of

and along the three coordinate axes.These equations are

Fnet,xMacom,xFnet,yMacom,yFnet,zMacom,z. (9-15)

Billiard Balls. Now we can go back and examine the behavior of the billiard

balls. Once the cue ball has begun to roll, no net external force acts on the (two-

ball) system. Thus, because 0, Eq. 9-14 tells us that 0 also. Because

acceleration is the rate of change of velocity, we conclude that the velocity of the

center of mass of the system of two balls does not change.When the two balls col-

lide, the forces that come into play are internal forces, on one ball from the other.

Such forces do not contribute to the net force , which remains zero. Thus, the

center of mass of the system, which was moving forward before the collision,

must continue to move forward after the collision, with the same speed and in the

same direction.

Solid Body. Equation 9-14 applies not only to a system of particles but also

to a solid body, such as the bat of Fig. 9-1b. In that case, Min Eq. 9-14 is the mass

of the bat and is the gravitational force on the bat. Equation 9-14 then tells us

that In other words, the center of mass of the bat moves as if the bat

were a single particle of mass M, with force acting on it.

Exploding Bodies. Figure 9-5 shows another interesting case. Suppose that at

a fireworks display, a rocket is launched on a parabolic path. At a certain point, it

explodes into fragments. If the explosion had not occurred, the rocket would have

continued along the trajectory shown in the figure. The forces of the explosion are

internal to the system (at first the system is just the rocket, and later it is its frag-

ments); that is, they are forces on parts of the system from other parts. If we ignore

air drag, the net external force acting on the system is the gravitational force on

the system, regardless of whether the rocket explodes. Thus, from Eq. 9-14, the ac-

celeration of the center of mass of the fragments (while they are in flight) re-

mains equal to This means that the center of mass of the fragments follows the

same parabolic trajectory that the rocket would have followed had it not exploded.

Ballet Leap. When a ballet dancer leaps across the stage in a grand jeté, she

raises her arms and stretches her legs out horizontally as soon as her feet leave the

com

net

com g

net

com

net

com

net

com

net

(Fnet

:ma

221

9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES

Figure 9-5 A fireworks rocket explodes in

flight. In the absence of air drag, the center

of mass of the fragments would continue to

follow the original parabolic path, until

fragments began to hit the ground.

The internal forces of the

explosion cannot change

the path of the com.

stage (Fig. 9-6). These actions shift her center of mass upward through her body.

Although the shifting center of mass faithfully follows a parabolic path across the

stage, its movement relative to the body decreases the height that is attained by her

head and torso, relative to that of a normal jump.The result is that the head and torso

follow a nearly horizontal path,giving an illusion that the dancer is floating.

Proof of Equation 9-14

Now let us prove this important equation. From Eq. 9-8 we have, for a system of n

particles,

(9-16)

in which Mis the system’s total mass and is the vector locating the position of

the system’s center of mass.

Differentiating Eq. 9-16 with respect to time gives

(9-17)

Here is the velocity of the ith particle, and is the

velocity of the center of mass.

Differentiating Eq. 9-17 with respect to time leads to

(9-18)

Here is the acceleration of the ith particle, and is

the acceleration of the center of mass. Although the center of mass is just a geo-

metrical point, it has a position, a velocity, and an acceleration, as if it were a particle.

From Newton’s second law, is equal to the resultant force that acts on

the ith particle.Thus, we can rewrite Eq. 9-18 as

(9-19)

Among the forces that contribute to the right side of Eq. 9-19 will be forces that

the particles of the system exert on each other (internal forces) and forces

exerted on the particles from outside the system (external forces). By Newton’s

third law, the internal forces form third-law force pairs and cancel out in the sum

that appears on the right side of Eq. 9-19. What remains is the vector sum of

all the external forces that act on the system. Equation 9-19 then reduces to

Eq. 9-14, the relation that we set out to prove.

com F1

:F2

:F3

:  Fn

miai

com (dv

:com /dt)a

i (dv

i/dt)

com m1a1

:m2a2

:m3a3

:  mnan

:com (dr

:com /dt)vi

: (dr

:/dt)

:com m1v1

:m2v2

:m3v3

:mnvn

rcom

com m1r1

:m2r2

:m3r3

:  mnrn

222 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Path of head

Path of center of mass

Figure 9-6 A grand jeté. (Based on The Physics of Dance, by Kenneth Laws, Schirmer

Books, 1984.)

223

9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES

Checkpoint 2

Two skaters on frictionless ice hold opposite ends of a pole of negligible mass.An axis

runs along it, with the origin at the center of mass of the two-skater system. One skater,

Fred, weighs twice as much as the other skater, Ethel.Where do the skaters meet if (a)

Fred pulls hand over hand along the pole so as to draw himself to Ethel, (b) Ethel pulls

hand over hand to draw herself to Fred, and (c) both skaters pull hand over hand?

Sample Problem 9.03 Motion of the com of three particles

If the particles in a system all move together, the com moves

with them—no trouble there. But what happens when they

move in different directions with different accelerations?

Here is an example.

The three particles in Fig. 9-7aare initially at rest. Each

experiences an external force due to bodies outside the

three-particle system. The directions are indicated, and the

magnitudes are F16.0 N, F212 N, and F314 N. What

is the acceleration of the center of mass of the system, and in

what direction does it move?

KEY IDEAS

The position of the center of mass is marked by a dot in the

figure. We can treat the center of mass as if it were a real

particle, with a mass equal to the system’s total mass M16 kg.

We can also treat the three external forces as if they act at the

center of mass (Fig.9-7b).

Calculations: We can now apply Newton’s second law

to the center of mass, writing

(9-20)

so (9-21)

Equation 9-20 tells us that the acceleration of the

center of mass is in the same direction as the net external force

on the system (Fig. 9-7b). Because the particles are ini-

tially at rest, the center of mass must also be at rest. As the

center of mass then begins to accelerate, it must move off in

the common direction of and

We can evaluate the right side of Eq. 9-21 directly on

a vector-capable calculator, or we can rewrite Eq. 9-21 in

component form, find the components of and then find

Along the xaxis, we have

6.0 N (12 N) cos 4514 N

16 kg 1.03 m/s2.

acom, xF1xF2xF3x

com.

com,

net.a

com

net

com

com F1

:F2

:F3

:F2

:F3

:Ma

com

net Ma

:com

net ma

Figure 9-7 (a) Three particles, initially at rest in the positions shown,

are acted on by the external forces shown.The center of mass (com)

of the system is marked. (b) The forces are now transferred to the

center of mass of the system, which behaves like a particle with a

mass Mequal to the total mass of the system.The net external force

and the acceleration of the center of mass are shown.

com

net

–1

–2

–3

–3 –2 –1 1 2 3 4 5

45°

8.0 kg

com

4.0 kg

com

M = 16 kg

(b)

(a)

F1F2

F2Fnet

acom

The com of the system

will move as if all the

mass were there and

the net force acted there.

Additional examples, video, and practice available at WileyPLUS

Along the yaxis, we have

From these components, we find that has the magnitude

(Answer)

and the angle (from the positive direction of the xaxis)

(Answer)



tan1acom, y

acom, x

27.

1.16 m/s21.2 m/s2

acom 2(acom, x)2(acom, y)2

com

0(12 N) sin 450

16 kg 0.530 m/s2.

acom, yF1yF2yF3y

Linear Momentum

Here we discuss only a single particle instead of a system of particles, in order to

define two important quantities. Then we shall extend those definitions to sys-

tems of many particles.

The first definition concerns a familiar word—momentum —that has several

meanings in everyday language but only a single precise meaning in physics and

engineering. The linear momentum of a particle is a vector quantity that is

defined as

(linear momentum of a particle), (9-22)

in which mis the mass of the particle and is its velocity. (The adjective linear is of-

ten dropped, but it serves to distinguish from angular momentum, which is intro-

duced in Chapter 11 and which is associated with rotation.) Since mis always a

positive scalar quantity, Eq. 9-22 tells us that and have the same direction. Fromv

:mv

224 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

9-3 LINEAR MOMENTUM

After reading this module, you should be able to . . .

9.13 Identify that momentum is a vector quantity and thus has

both magnitude and direction and also components.

9.14 Calculate the (linear) momentum of a particle as the

product of the particle’s mass and velocity.

9.15 Calculate the change in momentum (magnitude and di-

rection) when a particle changes its speed and direction of

travel.

9.16 Apply the relationship between a particle’s momentum

and the (net) force acting on the particle.

9.17 Calculate the momentum of a system of particles as the

product of the system’s total mass and its center-of-mass

velocity.

9.18 Apply the relationship between a system’s center-of-

mass momentum and the net force acting on the system.

●For a single particle, we define a quantity called its linear

momentum as

which is a vector quantity that has the same direction as the

particle’s velocity. We can write Newton’s second law in

:mv

:terms of this momentum:

●For a system of particles these relations become

and F

net dP

dt .P

:Mv

com

net dp

dt .

Learning Objectives

Key Ideas

The time rate of change of the momentum of a particle is equal to the net force

acting on the particle and is in the direction of that force.

In equation form this becomes

(9-23)

In words, Eq. 9-23 says that the net external force on a particle changes the

particle’s linear momentum Conversely, the linear momentum can be

changed only by a net external force. If there is no net external force, cannot

change.As we shall see in Module 9-5, this last fact can be an extremely power-

ful tool in solving problems.

net

net dp

dt .

Eq.9-22, the SI unit for momentum is the kilogram-meter per second (kg m/s).

Force and Momentum. Newton expressed his second law of motion in terms

of momentum:

Manipulating Eq. 9-23 by substituting for from Eq. 9-22 gives, for constant

mass m,

Thus, the relations and are equivalent expressions of

Newton’s second law of motion for a particle.

net ma

net dp

:/dt

net dp

dt d

dt (mv

:)mdv

dt ma

225

9-3 LINEAR MOMENTUM

Checkpoint 3

The figure gives the magnitude pof the linear mo-

mentum versus time tfor a particle moving along

an axis.A force directed along the axis acts on the

particle.(a) Rank the four regions indicated ac-

cording to the magnitude of the force,greatest

first.(b) In which region is the particle slowing?

The Linear Momentum of a System of Particles

Let’s extend the definition of linear momentum to a system of particles. Consider

a system of nparticles, each with its own mass, velocity, and linear momentum.

The particles may interact with each other, and external forces may act on them.

The system as a whole has a total linear momentum which is defined to be the

vector sum of the individual particles’ linear momenta.Thus,

(9-24)

If we compare this equation with Eq. 9-17, we see that

(linear momentum, system of particles), (9-25)

which is another way to define the linear momentum of a system of particles:

:Mv

:com

m1v

:1m2v

:2m3v

:3  mnv

:p

:1p

:2p

:3  p

The linear momentum of a system of particles is equal to the product of the total

mass Mof the system and the velocity of the center of mass.

Force and Momentum. If we take the time derivative of Eq. 9-25 (the veloc-

ity can change but not the mass), we find

(9-26)

Comparing Eqs. 9-14 and 9-26 allows us to write Newton’s second law for a sys-

tem of particles in the equivalent form

(system of particles), (9-27)

where is the net external force acting on the system.This equation is the gen-

eralization of the single-particle equation to a system of many

particles. In words, the equation says that the net external force on a system

of particles changes the linear momentum of the system. Conversely, the linear

momentum can be changed only by a net external force. If there is no net exter-

nal force, cannot change. Again, this fact gives us an extremely powerful tool

for solving problems.

net

net dp

:/dt

net

net dP

dt Mdv

:com

dt Ma

:com.

226 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

9-4 COLLISION AND IMPULSE

After reading this module, you should be able to . . .

9.19 Identify that impulse is a vector quantity and thus has both

magnitude and direction and also components.

9.20 Apply the relationship between impulse and momentum

change.

9.21 Apply the relationship between impulse, average force,

and the time interval taken by the impulse.

9.22 Apply the constant-acceleration equations to relate im-

pulse to average force.

9.23 Given force as a function of time, calculate the impulse (and

thus also the momentum change) by integrating the function.

9.24 Given a graph of force versus time, calculate the im-

pulse (and thus also the momentum change) by graphical

integration.

9.25 In a continuous series of collisions by projectiles, calcu-

late the average force on the target by relating it to the rate

at which mass collides and to the velocity change experi-

enced by each projectile.

●Applying Newton’s second law in momentum form to a

particle-like body involved in a collision leads to the

impulse–linear momentum theorem:

where is the change in the body’s linear momen-p

fp

ip

fp

ip

:J

●When a steady stream of bodies, each with mass mand

speed v, collides with a body whose position is fixed, the aver-

age force on the fixed body is

where n/tis the rate at which the bodies collide with the

fixed body, and vis the change in velocity of each colliding

body. This average force can also be written as

where m/tis the rate at which mass collides with the fixed

body. The change in velocity is vvif the bodies stop

upon impact and v2vif they bounce directly backward

with no change in their speed.

Favg m

tv,

Favg  n

tp n

tmv,

Learning Objectives

Key Ideas

tum, and is the impulse due to the force exerted on the

body by the other body in the collision:

●If Favg is the average magnitude of during the collisionF

(t)

:tf

(t)dt.

(t)J

and tis the duration of the collision, then for one-dimensional

motion

JFavg t.

Collision and Impulse

The momentum of any particle-like body cannot change unless a net

external force changes it. For example, we could push on the body to change its

momentum. More dramatically, we could arrange for the body to collide with a

baseball bat. In such a collision (or crash), the external force on the body is brief,

has large magnitude, and suddenly changes the body’s momentum. Collisions oc-

cur commonly in our world, but before we get to them, we need to consider a sim-

ple collision in which a moving particle-like body (a projectile) collides with some

other body (a target).

Single Collision

Let the projectile be a ball and the target be a bat.The collision is brief, and the ball

experiences a force that is great enough to slow, stop, or even reverse its motion.

Figure 9-8 depicts the collision at one instant.The ball experiences a force that

varies during the collision and changes the linear momentum of the ball. Thatp

(t)

The collision of a ball with a bat collapses

part of the ball.

Photo by Harold E. Edgerton. © The Harold and Esther Edgerton

Family Trust, courtesy of Palm Press, Inc.

change is related to the force by Newton’s second law written in the form

By rearranging this second-law expression, we see that, in time interval dt, the

change in the ball’s momentum is

(9-28)dp

:F

(t)dt.

:dp

:/dt.

We can find the net change in the ball’s momentum due to the collision if we inte-

grate both sides of Eq. 9-28 from a time tijust before the collision to a time tfjust

after the collision:

(9-29)

The left side of this equation gives us the change in momentum:

The right side, which is a measure of both the magnitude and the duration of the

collision force, is called the impulse of the collision:

(impulse defined). (9-30)

Thus, the change in an object’s momentum is equal to the impulse on the object:

(linear momentum–impulse theorem). (9-31)

This expression can also be written in the vector form

(9-32)

and in such component forms as

pxJx(9-33)

and (9-34)

Integrating the Force. If we have a function for we can evaluate (and

thus the change in momentum) by integrating the function. If we have a plot of

versus time t, we can evaluate by finding the area between the curve and the t

axis, such as in Fig. 9-9a. In many situations we do not know how the force varies

with time but we do know the average magnitude Favg of the force and the duration

t(tfti) of the collision. Then we can write the magnitude of the impulse as

JFavg t. (9-35)

The average force is plotted versus time as in Fig. 9-9b.The area under that curve

is equal to the area under the curve for the actual force F(t) in Fig. 9-9abecause

both areas are equal to impulse magnitude J.

Instead of the ball, we could have focused on the bat in Fig. 9-8. At any

instant, Newton’s third law tells us that the force on the bat has the same

magnitude but the opposite direction as the force on the ball. From Eq. 9-30, this

means that the impulse on the bat has the same magnitude but the opposite

direction as the impulse on the ball.

:(t),

pfx pix tf

Fxdt.

fp

iJ

p

:J

:tf

(t)dt

fp

ip

tf

:tf

(t)dt.

227

9-4 COLLISION AND IMPULSE

Figure 9-8 Force acts on a ball as the

ball and a bat collide.

:(t)

Bat Ball

F(t)

Figure 9-9 (a) The curve shows the magni-

tude of the time-varying force F(t) that acts

on the ball in the collision of Fig. 9-8. The

area under the curve is equal to the magni-

tude of the impulse on the ball in the col-

lision. (b) The height of the rectangle repre-

sents the average force Favg acting on the

ball over the time interval t.The area within

the rectangle is equal to the area under the

curve in (a) and thus is also equal to the

magnitude of the impulse in the collision.J

F(t)

Δt

Favg

(a)

(b)

The impulse in the collision

is equal to the area under

the curve.

The average force gives

the same area under the

curve.

Checkpoint 4

A paratrooper whose chute fails to open lands in snow; he is hurt slightly. Had he

landed on bare ground, the stopping time would have been 10 times shorter and the

collision lethal. Does the presence of the snow increase,decrease, or leave unchanged

the values of (a) the paratrooper’s change in momentum,(b) the impulse stopping the

paratrooper,and (c) the force stopping the paratrooper?

Series of Collisions

Now let’s consider the force on a body when it undergoes a series of identical, re-

peated collisions. For example, as a prank, we might adjust one of those machines

that fire tennis balls to fire them at a rapid rate directly at a wall. Each collision

would produce a force on the wall, but that is not the force we are seeking. We

want the average force Favg on the wall during the bombardment—that is, the av-

erage force during a large number of collisions.

In Fig. 9-10, a steady stream of projectile bodies, with identical mass mand

linear momenta moves along an xaxis and collides with a target body that ismv

228 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Figure 9-10 A steady stream of projectiles,

with identical linear momenta, collides

with a target, which is fixed in place. The

average force Favg on the target is to the

right and has a magnitude that depends on

the rate at which the projectiles collide

with the target or, equivalently, the rate at

which mass collides with the target.

Target

Projectiles

fixed in place. Let nbe the number of projectiles that collide in a time interval t.

Because the motion is along only the xaxis, we can use the components of the

momenta along that axis. Thus, each projectile has initial momentum mv and

undergoes a change pin linear momentum because of the collision. The total

change in linear momentum for nprojectiles during interval tis np.The

resulting impulse on the target during tis along the xaxis and has the same

magnitude of npbut is in the opposite direction. We can write this relation in

component form as

Jnp, (9-36)

where the minus sign indicates that Jand phave opposite directions.

Average Force. By rearranging Eq. 9-35 and substituting Eq. 9-36, we find

the average force Favg acting on the target during the collisions:

(9-37)

This equation gives us Favg in terms of n/t, the rate at which the projectiles

collide with the target, and v, the change in the velocity of those projectiles.

Velocity Change. If the projectiles stop upon impact, then in Eq. 9-37 we can

substitute, for v,

vvfvi0vv, (9-38)

where vi(v) and vf(0) are the velocities before and after the collision,

respectively. If, instead, the projectiles bounce (rebound) directly backward from

the target with no change in speed, then vfvand we can substitute

vvfvivv2v. (9-39)

In time interval t, an amount of mass mnm collides with the target.

With this result, we can rewrite Eq. 9-37 as

(9-40)

This equation gives the average force Favg in terms of m/t, the rate at which

mass collides with the target. Here again we can substitute for vfrom Eq. 9-38

or 9-39 depending on what the projectiles do.

Favg m

tv.

Favg J

t n

tp n

tmv.

Checkpoint 5

The figure shows an overhead view of a ball bouncing from a vertical wall without any

change in its speed. Consider the change in the ball’s linear momentum.(a) Is px

positive, negative, or zero? (b) Is pypositive,negative, or zero? (c) What is the direc-

tion of ?p

p

229

9-4 COLLISION AND IMPULSE

Impulse: The impulse is then

(Answer)

which means the impulse magnitude is

The angle of is given by

(Answer)

which a calculator evaluates as 75.4. Recall that the physi-

cally correct result of an inverse tangent might be the

displayed answer plus 180.We can tell which is correct here

by drawing the components of (Fig. 9-11c). We find that u

is actually 75.4180255.4, which we can write as

u105. (Answer)

(b) The collision lasts for 14 ms. What is the magnitude of

the average force on the driver during the collision?

KEY IDEA

From Eq. 9-35 (J Favg t), the magnitude Favg of the aver-

age force is the ratio of the impulse magnitude Jto the dura-

tion tof the collision.

Calculations: We have

. (Answer)

Using Fma with m80 kg, you can show that the magni-

tude of the driver’s average acceleration during the collision

is about 3.22 103m/s2329g, which is fatal.

Surviving: Mechanical engineers attempt to reduce the

chances of a fatality by designing and building racetrack

walls with more “give,” so that a collision lasts longer. For

example, if the collision here lasted 10 times longer and the

other data remained the same, the magnitudes of the aver-

age force and average acceleration would be 10 times less

and probably survivable.

2.583 105 N 2.6 105 N

Favg J

t3616 kgm/s

0.014 s

utan1Jy

J2Jx

2Jy

23616 kgm/s 3600 kgm/s.

:(910i

ˆ3500j

ˆ) kgm/s,

Sample Problem 9.04 Two-dimensional impulse, race car–wall collision

Figure 9-11ais an overhead view of

the path taken by a race car driver as his car collides with the

racetrack wall. Just before the collision, he is traveling at

speed vi70 m/s along a straight line at 30from the wall.

Just after the collision, he is traveling at speed vf50 m/s

along a straight line at 10from the wall. His mass mis 80 kg.

(a) What is the impulse on the driver due to the collision?

KEY IDEAS

We can treat the driver as a particle-like body and thus apply

the physics of this module. However, we cannot calculate

directly from Eq. 9-30 because we do not know anything about

the force on the driver during the collision. That is, we do

not have a function of or a plot for it and thus cannot

integrate to find . However,we can find from the change in

the driver’s linear momentum via Eq. 9-32 .

Calculations: Figure 9-11bshows the driver’s momentum p

:p

fp

i)p

:(t)

Race car–wall collision.

Wall

30°

10°

30°

Path

(a)

10°

(b)

pf–105°

(c)

The impulse on the car

is equal to the change

in the momentum.

The collision

changes the

momentum.

Figure 9-11 (a) Overhead

view of the path taken by a

race car and its driver as the

car slams into the racetrack

wall. (b) The initial momen-

tum and final momentum

of the driver. (c) The

impulse on the driver

during the collision.

Additional examples, video, and practice available at WileyPLUS

before the collision (at angle 30from the positive xdirection)

and his momentum after the collision (at angle 10). From

Eqs. 9-32 and 9-22 ,we can write

(9-41)

We could evaluate the right side of this equation directly on

a vector-capable calculator because we know mis 80 kg,

is 50 m/s at 10, and is 70 m/s at 30. Instead, here we

evaluate Eq. 9-41 in component form.

x component: Along the xaxis we have

Jxm(vfx vix)

(80 kg)[(50 m/s) cos(10)(70 m/s) cos 30]

910 kgm/s.

y component: Along the yaxis,

Jym(vfy viy)

(80 kg)[(50 m/s) sin(10)(70 m/s) sin 30]

3495 kgm/s 3500 kgm/s.

:p

fp

imv

fmvi

:m(v

fv

i).

:)( p

:

p

Conservation of Linear Momentum

Suppose that the net external force (and thus the net impulse ) acting on a

system of particles is zero (the system is isolated) and that no particles leave or

enter the system (the system is closed). Putting in Eq. 9-27 then yields

, which means that

(closed, isolated system). (9-42)

In words,

:constant

:/dt 0

net 0

net

230 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

9-5 CONSERVATION OF LINEAR MOMENTUM

After reading this module, you should be able to . . .

9.26 For an isolated system of particles, apply the conservation

of linear momenta to relate the initial momenta of the particles

to their momenta at a later instant.

9.27 Identify that the conservation of linear momentum can be

done along an individual axis by using components along

that axis, provided that there is no net external force com-

ponent along that axis.

●If a system is closed and isolated so that no net external

force acts on it, then the linear momentum must be constant

even if there are internal changes:

(closed, isolated system).P

:constant

●This conservation of linear momentum can also be written

in terms of the system’s initial momentum and its momentum

at some later instant:

(closed, isolated system),P

iP

Learning Objectives

Key Ideas

If no net external force acts on a system of particles, the total linear momentum

of the system cannot change.

This result is called the law of conservation of linear momentum and is an extremely

powerful tool in solving problems. In the homework we usually write the law as

(closed, isolated system). (9-43)

In words, this equation says that, for a closed, isolated system,

Caution: Momentum should not be confused with energy. In the sample problems

of this module, momentum is conserved but energy is definitely not.

Equations 9-42 and 9-43 are vector equations and, as such, each is equivalent

to three equations corresponding to the conservation of linear momentum in

three mutually perpendicular directions as in, say, an xyz coordinate system.

Depending on the forces acting on a system, linear momentum might be

conserved in one or two directions but not in all directions. However,



total linear momentum

at some initial time ti







total linear momentum

at some later time tf



iP

If the component of the net external force on a closed system is zero along an axis, then

the component of the linear momentum of the system along that axis cannot change.

In a homework problem, how can you know if linear momentum can be con-

served along,say, an xaxis? Check the force components along that axis. If the net of

any such components is zero, then the conservation applies.As an example, suppose

that you toss a grapefruit across a room. During its flight, the only external force act-

ing on the grapefruit (which we take as the system) is the gravitational force ,

which is directed vertically downward. Thus, the vertical component of the linear

momentum of the grapefruit changes, but since no horizontal external force acts on

the grapefruit,the horizontal component of the linear momentum cannot change.

Note that we focus on the external forces acting on a closed system.

Although internal forces can change the linear momentum of portions of the sys-

tem, they cannot change the total linear momentum of the entire system. For ex-

ample, there are plenty of forces acting between the organs of your body, but they

do not propel you across the room (thankfully).

The sample problems in this module involve explosions that are either one-

dimensional (meaning that the motions before and after the explosion are along

a single axis) or two-dimensional (meaning that they are in a plane containing

two axes). In the following modules we consider collisions.

231

9-5 CONSERVATION OF LINEAR MOMENTUM

We can relate the vMS to the known velocities with

In symbols, this gives us

vHS vrel vMS (9-47)

or vMS vHS vrel.

Substituting this expression for vMS into Eq. 9-46, and then

substituting Eqs. 9-45 and 9-46 into Eq. 9-44,we find

Mvi0.20M(vHS vrel)0.80MvHS,

which gives us

vHS vi0.20vrel,

or vHS 2100 km/h (0.20)(500 km/h)

2200 km/h. (Answer)



velocity of

hauler relative

to Sun







velocity of

hauler relative

to module







velocity of

module relative

to Sun



Sample Problem 9.05 One-dimensional explosion, relative velocity, space hauler

One-dimensional explosion: Figure 9-12ashows a space hauler

and cargo module, of total mass M, traveling along an xaxis in

deep space. They have an initial velocity of magnitude 2100

km/h relative to the Sun. With a small explosion, the hauler

ejects the cargo module, of mass 0.20M(Fig. 9-12b).The hauler

then travels 500 km/h faster than the module along the xaxis;

that is, the relative speed vrel between the hauler and the mod-

ule is 500 km/h.What then is the velocity of the hauler rela-

tive to the Sun?

KEY IDEA

Because the hauler–module system is closed and isolated,

its total linear momentum is conserved; that is,

, (9-44)

where the subscripts iand frefer to values before and after

the ejection, respectively. (We need to be careful here:

Although the momentum of the system does not change, the

momenta of the hauler and module certainly do.)

Calculations: Because the motion is along a single axis, we can

write momenta and velocities in terms of their xcomponents,

using a sign to indicate direction. Before the ejection, we have

PiMvi. (9-45)

Let vMS be the velocity of the ejected module relative to the

Sun.The total linear momentum of the system after the ejec-

tion is then

Pf(0.20M)vMS (0.80M)vHS, (9-46)

where the first term on the right is the linear momentum of

the module and the second term is that of the hauler.

iP

Figure 9-12 (a) A space hauler, with a cargo module, moving at initial

velocity (b) The hauler has ejected the cargo module. Now the

velocities relative to the Sun are for the module and for the

hauler.

(

)(

)

Cargo module

Hauler

0.20M

vMS vHS

0.80M

The explosive separation can change the momentum

of the parts but not the momentum of the system.

Additional examples, video, and practice available at WileyPLUS

Checkpoint 6

An initially stationary device lying on a frictionless floor explodes into two pieces,which

then slide across the floor,one of them in the positive xdirection. (a) What is the sum of

the momenta of the two pieces after the explosion? (b) Can the second piece move at an

angle to the xaxis? (c) What is the direction of the momentum of the second piece?

232 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Calculations: Linear momentum is also conserved along

the xaxis because there is no net external force acting on

the coconut and pieces along that axis.Thus we have

Pix Pfx, (9-49)

where Pix 0 because the coconut is initially at rest. To

get Pfx, we find the xcomponents of the final momenta,

using the fact that piece Amust have a mass of 0.50M

(M0.20M0.30M):

pfA,x 0.50MvfA,

pfB,x 0.20MvfB,x 0.20MvfB cos 50,

pfC,x 0.30MvfC,x 0.30MvfC cos 80.

Equation 9-49 for the conservation of momentum along the

xaxis can now be written as

Pix Pfx pfA,x pfB,x pfC,x.

Then, with vfC 5.0 m/s and vfB 9.64 m/s, we have

00.50MvfA 0.20M(9.64 m/s) cos 50

0.30M(5.0 m/s) cos 80,

from which we find

vfA 3.0 m/s. (Answer)

Sample Problem 9.06 Two-dimensional explosion, momentum, coconut

Two-dimensional explosion: A firecracker placed inside a

coconut of mass M, initially at rest on a frictionless floor,

blows the coconut into three pieces that slide across the floor.

An overhead view is shown in Fig. 9-13a. Piece C, with mass

0.30M, has final speed vfC 5.0 m/s.

(a) What is the speed of piece B, with mass 0.20M?

KEY IDEA

First we need to see whether linear momentum is con-

served. We note that (1) the coconut and its pieces form a

closed system, (2) the explosion forces are internal to that

system, and (3) no net external force acts on the system.

Therefore, the linear momentum of the system is conserved.

(We need to be careful here: Although the momentum of

the system does not change, the momenta of the pieces cer-

tainly do.)

Calculations: To get started, we superimpose an xy coordinate

system as shown in Fig. 9-13b, with the negative direction of the

xaxis coinciding with the direction of The xaxis is at 80v

fA.

Additional examples, video, and practice available at WileyPLUS

Figure 9-13 Three pieces of an

exploded coconut move off in

three directions along a

frictionless floor. (a) An over-

head view of the event. (b) The

same with a two-dimensional

axis system imposed.

with the direction of and 50with the direction of .

Linear momentum is conserved separately along each

axis. Let’s use the yaxis and write

Piy Pfy, (9-48)

where subscript irefers to the initial value (before the ex-

plosion), and subscript yrefers to the ycomponent of

or .

The component Piy of the initial linear momentum is

zero, because the coconut is initially at rest. To get an ex-

pression for Pfy, we find the ycomponent of the final linear

momentum of each piece, using the y-component version of

Eq. 9-22 ( pymvy):

pfA,y 0,

pfB,y 0.20MvfB,y 0.20MvfB sin 50,

pfC,y 0.30MvfC,y 0.30MvfC sin 80.

(Note that pfA,y 0 because of our nice choice of axes.)

Equation 9-48 can now be written as

Piy Pfy pfA,y pfB,y pfC,y.

Then, with vfC 5.0 m/s, we have

000.20MvfB sin 50(0.30M)(5.0 m/s) sin 80,

from which we find

vfB 9.64 m/s 9.6 m/s. (Answer)

(b) What is the speed of piece A?

:Pi

vfB

vfC

vfA

100°

130°

(a)

vfB

vfC

vfA

80°

(b)

50°

The explosive separation

can change the momentum

of the parts but not the

momentum of the system.

Momentum and Kinetic Energy in Collisions

In Module 9-4, we considered the collision of two particle-like bodies but focused

on only one of the bodies at a time. For the next several modules we switch our

focus to the system itself, with the assumption that the system is closed and iso-

lated. In Module 9-5, we discussed a rule about such a system: The total linear

momentum of the system cannot change because there is no net external force

to change it.This is a very powerful rule because it can allow us to determine the

results of a collision without knowing the details of the collision (such as how

much damage is done).

We shall also be interested in the total kinetic energy of a system of two col-

liding bodies. If that total happens to be unchanged by the collision, then the

kinetic energy of the system is conserved (it is the same before and after the

collision). Such a collision is called an elastic collision. In everyday collisions of

common bodies, such as two cars or a ball and a bat, some energy is always trans-

ferred from kinetic energy to other forms of energy, such as thermal energy or

energy of sound. Thus, the kinetic energy of the system is not conserved. Such a

collision is called an inelastic collision.

However, in some situations, we can approximate a collision of common bod-

ies as elastic. Suppose that you drop a Superball onto a hard floor. If the collision

between the ball and floor (or Earth) were elastic, the ball would lose no kinetic

energy because of the collision and would rebound to its original height.

However, the actual rebound height is somewhat short, showing that at least

some kinetic energy is lost in the collision and thus that the collision is somewhat

inelastic. Still, we might choose to neglect that small loss of kinetic energy to ap-

proximate the collision as elastic.

The inelastic collision of two bodies always involves a loss in the kinetic

energy of the system. The greatest loss occurs if the bodies stick together, in

which case the collision is called a completely inelastic collision. The collision of a

baseball and a bat is inelastic. However, the collision of a wet putty ball and a bat

is completely inelastic because the putty sticks to the bat.

233

9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS

After reading this module, you should be able to . . .

9.28 Distinguish between elastic collisions, inelastic collisions,

and completely inelastic collisions.

9.29 Identify a one-dimensional collision as one where the ob-

jects move along a single axis, both before and after the

collision.

9.30 Apply the conservation of momentum for an isolated

one-dimensional collision to relate the initial momenta of

the objects to their momenta after the collision.

9.31 Identify that in an isolated system, the momentum and

velocity of the center of mass are not changed even if the

objects collide.

●In an inelastic collision of two bodies, the kinetic energy of

the two-body system is not conserved. If the system is closed

and isolated, the total linear momentum of the system must

be conserved, which we can write in vector form as

where subscripts iand frefer to values just before and just

after the collision, respectively.

●If the motion of the bodies is along a single axis, the collision

is one-dimensional and we can write the equation in terms of

:1ip

:2ip

:1fp

:2f

velocity components along that axis:

m1v1im2v2im1v1fm2v2f.

●If the bodies stick together, the collision is a completely

inelastic collision and the bodies have the same final veloc-

ity V(because they are stuck together).

●The center of mass of a closed, isolated system of two col-

liding bodies is not affected by a collision. In particular, the ve-

locity of the center of mass cannot be changed by the

collision.

:com

Learning Objectives

Key Ideas

Inelastic Collisions in One Dimension

One-Dimensional Inelastic Collision

Figure 9-14 shows two bodies just before and just after they have a one-

dimensional collision. The velocities before the collision (subscript i) and after

the collision (subscript f) are indicated.The two bodies form our system, which is

closed and isolated.We can write the law of conservation of linear momentum for

this two-body system as

which we can symbolize as

(conservation of linear momentum). (9-50)

Because the motion is one-dimensional, we can drop the overhead arrows for

vectors and use only components along the axis, indicating direction with a sign.

Thus, from pmv, we can rewrite Eq. 9-50 as

m1v1im2v2im1v1fm2v2f. (9-51)

If we know values for,say, the masses, the initial velocities, and one of the final ve-

locities, we can find the other final velocity with Eq. 9-51.

One-Dimensional Completely Inelastic Collision

Figure 9-15 shows two bodies before and after they have a completely inelastic

collision (meaning they stick together).The body with mass m2happens to be ini-

tially at rest (v2i0).We can refer to that body as the target and to the incoming

body as the projectile. After the collision, the stuck-together bodies move with

velocity V. For this situation, we can rewrite Eq. 9-51 as

m1v1i(m1m2)V(9-52)

or . (9-53)

If we know values for, say, the masses and the initial velocity v1iof the projectile,

we can find the final velocity Vwith Eq. 9-53. Note that Vmust be less than v1ibe-

cause the mass ratio m1/(m1m2) must be less than unity.

Velocity of the Center of Mass

In a closed, isolated system, the velocity of the center of mass of the system

cannot be changed by a collision because, with the system isolated, there is no net

external force to change it. To get an expression for , let us return to the v

com

Vm1

m1m2

v1i

1ip

2ip

1fp



total momentum P

before the collision







total momentum P

after the collision



234 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Figure 9-14 Bodies 1 and 2 move along an

xaxis, before and after they have an

inelastic collision.

m1m2

Before

Body 1 Body 2

v1iv2i

m1m2

After

v1fv2f

Here is the generic setup

for an inelastic collision.

Figure 9-15 A completely inelastic collision between

two bodies. Before the collision, the body with mass

m2is at rest and the body with mass m1moves

directly toward it. After the collision, the stuck-

together bodies move with the same velocity .V

Projectile

Target

v1i

After

Before

m1 + m2

v2i = 0

In a completely inelastic

collision, the bodies

stick together.

two-body system and one-dimensional collision of Fig. 9-14. From Eq. 9-25

, we can relate to the total linear momentum of that two-body

system by writing

. (9-54)

The total linear momentum is conserved during the collision; so it is given by

either side of Eq. 9-50. Let us use the left side to write

. (9-55)

Substituting this expression for in Eq. 9-54 and solving for give us

. (9-56)

The right side of this equation is a constant, and has that same constant value

before and after the collision.

For example, Fig. 9-16 shows, in a series of freeze-frames, the motion of the

center of mass for the completely inelastic collision of Fig. 9-15. Body 2 is the tar-

get, and its initial linear momentum in Eq. 9-56 is Body 1 is

the projectile, and its initial linear momentum in Eq. 9-56 is Note

that as the series of freeze-frames progresses to and then beyond the collision,

the center of mass moves at a constant velocity to the right. After the

collision, the common final speed Vof the bodies is equal to because then

the center of mass travels with the stuck-together bodies.

com

1im1v

1i.

2im2v

2i0.

com

com P

m1m2

p

1ip

m1m2

com

:p

:1ip

:2i

:Mv

com (m1m2)v

com

:Mv

com)

235

9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS

v1iv2i= 0

m1+m2

V=vcom

Collision!

vcom

The com of the two

bodies is between

them and moves at a

constant velocity.

Here is the

incoming projectile.

The com moves at the

same velocity even after

the bodies stick together.

Here is the

stationary target.

Figure 9-16 Some freeze-frames of the two-body system

in Fig. 9-15, which undergoes a completely inelastic col-

lision. The system’s center of mass is shown in each

freeze-frame. The velocity of the center of mass is

unaffected by the collision. Because the bodies stick

together after the collision, their common velocity

must be equal to .v

com

Checkpoint 7

Body 1 and body 2 are in a completely inelastic one-dimensional collision.What is

their final momentum if their initial momenta are, respectively, (a) 10 kgm/s and 0;

(b) 10 kgm/s and 4 kgm/s;(c) 10 kgm/s and 4kgm/s?

236 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Additional examples, video, and practice available at WileyPLUS

There are two events here.

The bullet collides with the

block. Then the bullet–block

system swings upward by

height h.

Figure 9-17 A ballistic

pendulum, used to

measure the speeds

of bullets.

Sample Problem 9.07 Conservation of momentum, ballistic pendulum

Here is an example of a common technique in physics. We

have a demonstration that cannot be worked out as a whole

(we don’t have a workable equation for it). So, we break it

up into steps that can be worked separately (we have equa-

tions for them).

The ballistic pendulum was used to measure the speeds of

bullets before electronic timing devices were developed. The

version shown in Fig. 9-17 consists of a large block of wood of

mass M5.4 kg, hanging from two long cords. A bullet of

mass m9.5 g is fired into the block, coming quickly to rest.

The block bullet then swing upward, their center of mass

rising a vertical distance h6.3 cm before the pendulum

comes momentarily to rest at the end of its arc. What is the

speed of the bullet just prior to the collision?

KEY IDEAS

We can see that the bullet’s speed vmust determine the rise

height h.However, we cannot use the conservation of mechan-

ical energy to relate these two quantities because surely energy

is transferred from mechanical energy to other forms (such as

thermal energy and energy to break apart the wood) as the

bullet penetrates the block. Nevertheless, we can split this com-

plicated motion into two steps that we can separately analyze:

(1) the bullet–block collision and (2) the bullet–block rise,

during which mechanical energy is conserved.

Reasoning step 1: Because the collision within the

bullet–block system is so brief, we can make two impor-

tant assumptions: (1) During the collision, the gravita-

tional force on the block and the force on the block from

the cords are still balanced. Thus, during the collision, the

net external impulse on the bullet–block system is zero.

Therefore, the system is isolated and its total linear momen-

tum is conserved:

(9-57)

(2) The collision is one-dimensional in the sense that the di-

rection of the bullet and block just after the collision is in the

bullet’s original direction of motion.

Because the collision is one-dimensional, the block is ini-

tially at rest,and the bullet sticks in the block, we use Eq.9-53 to

express the conservation of linear momentum. Replacing the

symbols there with the corresponding symbols here,we have

(9-58)

Reasoning step 2: As the bullet and block now swing up to-

gether, the mechanical energy of the bullet–block–Earth

Vm

mMv.



total momentum

before the collision







total momentum

after the collision



system is conserved:

(9-59)

(This mechanical energy is not changed by the force of the

cords on the block, because that force is always directed

perpendicular to the block’s direction of travel.) Let’s take the

block’s initial level as our reference level of zero gravitational

potential energy. Then conservation of mechanical energy

means that the system’s kinetic energy at the start of the swing

must equal its gravitational potential energy at the highest

point of the swing. Because the speed of the bullet and block

at the start of the swing is the speed Vimmediately after the

collision,we may write this conservation as

(9-60)

Combining steps: Substituting for Vfrom Eq. 9-58 leads to

(9-61)

(Answer)

The ballistic pendulum is a kind of “transformer,” exchang-

ing the high speed of a light object (the bullet) for the low—

and thus more easily measurable—speed of a massive ob-

ject (the block).

630 m/s.





0.0095 kg 5.4 kg

0.0095 kg



2(2)(9.8 m/s2)(0.063 m)

vmM

m22gh

2(mM)V2(mM)gh.



mechanical energy

at bottom







mechanical energy

at top



Elastic Collisions in One Dimension

As we discussed in Module 9-6, everyday collisions are inelastic but we can

approximate some of them as being elastic; that is, we can approximate that the

total kinetic energy of the colliding bodies is conserved and is not transferred to

other forms of energy:

. (9-62)

This means:



total kinetic energy

before the collision







total kinetic energy

after the collision



237

9-7 ELASTIC COLLISIONS IN ONE DIMENSION

After reading this module, you should be able to . . .

9.32 For isolated elastic collisions in one dimension, apply the

conservation laws for both the total energy and the net mo-

mentum of the colliding bodies to relate the initial values to

the values after the collision.

9.33 For a projectile hitting a stationary target, identify the re-

sulting motion for the three general cases: equal masses,

target more massive than projectile, projectile more mas-

sive than target.

●An elastic collision is a special type of collision in which

the kinetic energy of a system of colliding bodies is con-

served. If the system is closed and isolated, its linear mo-

mentum is also conserved. For a one-dimensional collision in

which body 2 is a target and body 1 is an incoming projec-

tile, conservation of kinetic energy and linear momentum

yield the following expressions for the velocities immediately

after the collision:

and v2f2m1

m1m2

v1i.

v1fm1m2

m1m2

v1i

Learning Objectives

Key Idea

In an elastic collision, the kinetic energy of each colliding body may change, but

the total kinetic energy of the system does not change.

For example, the collision of a cue ball with an object ball in a game of pool

can be approximated as being an elastic collision. If the collision is head-on

(the cue ball heads directly toward the object ball), the kinetic energy of the cue

ball can be transferred almost entirely to the object ball. (Still, the collision trans-

fers some of the energy to the sound you hear.)

Stationary Target

Figure 9-18 shows two bodies before and after they have a one-dimensional colli-

sion, like a head-on collision between pool balls. A projectile body of mass m1

and initial velocity v1imoves toward a target body of mass m2that is initially at

rest (v2i0). Let’s assume that this two-body system is closed and isolated.Then

the net linear momentum of the system is conserved, and from Eq. 9-51 we can write

that conservation as

m1v1im1v1fm2v2f(linear momentum). (9-63)

If the collision is also elastic, then the total kinetic energy is conserved and we

can write that conservation as

(kinetic energy). (9-64)

In each of these equations, the subscript iidentifies the initial velocities and the

subscript fthe final velocities of the bodies. If we know the masses of the bodies

and if we also know v1i, the initial velocity of body 1, the only unknown quantities

are v1fand v2f, the final velocities of the two bodies.With two equations at our dis-

posal, we should be able to find these two unknowns.

2m1v1i

21

2m1v1f

21

2m2v2f

Figure 9-18 Body 1 moves along an xaxis

before having an elastic collision with

body 2, which is initially at rest. Both

bodies move along that axis after the

collision.

Before v1i

Projectile

Target

v2i = 0

After

v1f

m1m2

v2f

Here is the generic setup

for an elastic collision with

a stationary target.

To do so, we rewrite Eq. 9-63 as

m1(v1iv1f)m2v2f(9-65)

and Eq. 9-64 as*

(9-66)

After dividing Eq. 9-66 by Eq. 9-65 and doing some more algebra, we obtain

(9-67)

and (9-68)

Note that v2fis always positive (the initially stationary target body with mass m2

always moves forward). From Eq. 9-67 we see that v1fmay be of either sign (the

projectile body with mass m1moves forward if m1m2but rebounds if m1m2).

Let us look at a few special situations.

1. Equal masses If m1m2, Eqs. 9-67 and 9-68 reduce to

v1f0 and v2fv1i,

which we might call a pool player’s result. It predicts that after a head-on colli-

sion of bodies with equal masses, body 1 (initially moving) stops dead in its

tracks and body 2 (initially at rest) takes off with the initial speed of body 1. In

head-on collisions, bodies of equal mass simply exchange velocities. This is

true even if body 2 is not initially at rest.

2. A massive target In Fig. 9-18, a massive target means that m2m1.For

example, we might fire a golf ball at a stationary cannonball. Equations 9-67

and 9-68 then reduce to

(9-69)

This tells us that body 1 (the golf ball) simply bounces back along its incom-

ing path, its speed essentially unchanged. Initially stationary body 2 (the

cannonball) moves forward at a low speed, because the quantity in paren-

theses in Eq. 9-69 is much less than unity.All this is what we should expect.

3. A massive projectile This is the opposite case; that is, m1m2. This time, we

fire a cannonball at a stationary golf ball.Equations 9-67 and 9-68 reduce to

v1fv1iand v2f2v1i. (9-70)

Equation 9-70 tells us that body 1 (the cannonball) simply keeps on going,

scarcely slowed by the collision. Body 2 (the golf ball) charges ahead at twice

the speed of the cannonball. Why twice the speed? Recall the collision de-

scribed by Eq. 9-69, in which the velocity of the incident light body (the golf

ball) changed from vto v, a velocity change of 2v.The same change in ve-

locity (but now from zero to 2v) occurs in this example also.

Moving Target

Now that we have examined the elastic collision of a projectile and a stationary

target, let us examine the situation in which both bodies are moving before they

undergo an elastic collision.

For the situation of Fig. 9-19, the conservation of linear momentum is written as

m1v1im2v2im1v1fm2v2f, (9-71)



v1fv1i and v2f



2m1



v1i.



v2f2m1

m1m2

v1i.

v1fm1m2

m1m2

v1i

m1(v1iv1f)(v1iv1f)m2v2f

238 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

*In this step, we use the identity a2b2(ab)(ab). It reduces the amount of algebra needed to

solve the simultaneous equations Eqs. 9-65 and 9-66.

and the conservation of kinetic energy is written as

(9-72)

To solve these simultaneous equations for v1fand v2f, we first rewrite Eq. 9-71 as

m1(v1iv1f)m2(v2iv2f), (9-73)

and Eq. 9-72 as

m1(v1iv1f)(v1iv1f)m2(v2iv2f)(v2iv2f). (9-74)

After dividing Eq. 9-74 by Eq. 9-73 and doing some more algebra, we obtain

(9-75)

and (9-76)

Note that the assignment of subscripts 1 and 2 to the bodies is arbitrary. If we ex-

change those subscripts in Fig. 9-19 and in Eqs. 9-75 and 9-76, we end up with the

same set of equations. Note also that if we set v2i0, body 2 becomes a stationary

target as in Fig.9-18, and Eqs.9-75 and 9-76 reduce to Eqs. 9-67 and 9-68, respectively.

v2f2m1

m1m2

v1im2m1

m1m2

v2i.

v1fm1m2

m1m2

v1i2m2

m1m2

v2i

2m1v1i

21

2m2v2i

21

2m1v1f

21

2m2v2f

239

9-7 ELASTIC COLLISIONS IN ONE DIMENSION

Figure 9-19 Two bodies headed for a one-

dimensional elastic collision.

v1i

v2i

Here is the generic setup

for an elastic collision with

a moving target.

Checkpoint 8

What is the final linear momentum of the target in Fig. 9-18 if the initial linear momen-

tum of the projectile is 6 kgm/s and the final linear momentum of the projectile is (a)

2kgm/s and (b) 2kgm/s? (c) What is the final kinetic energy of the target if the

initial and final kinetic energies of the projectile are, respectively, 5 J and 2 J?

two reasons, we can apply Eqs. 9-67 and 9-68 to each of the

collisions.

Calculations: If we start with the first collision, we have too

many unknowns to make any progress: we do not know the

masses or the final velocities of the blocks. So, let’s start with

the second collision in which block 2 stops because of its col-

lision with block 3. Applying Eq. 9-67 to this collision, with

changes in notation, we have

where v2iis the velocity of block 2 just before the collision

and v2fis the velocity just afterward. Substituting v2f0

(block 2 stops) and then m36.0 kg gives us

(Answer)

With similar notation changes, we can rewrite Eq. 9-68 for

the second collision as

where v3fis the final velocity of block 3. Substituting m2m3

and the given v3f5.0 m/s, we find

v2iv3f5.0 m/s.

v3f2m2

m2m3

v2i,

m2m36.00 kg.

v2fm2m3

m2m3

v2i,

Sample Problem 9.08 Chain reaction of elastic collisions

Figure 9-20 Block 1 collides with stationary block 2, which then

collides with stationary block 3.

In Fig. 9-20a, block 1 approaches a line of two stationary

blocks with a velocity of v1i10 m/s. It collides with block 2,

which then collides with block 3, which has mass m36.0 kg.

After the second collision, block 2 is again stationary and

block 3 has velocity v3f 5.0 m/s (Fig.9-20b).Assume that the

collisions are elastic. What are the masses of blocks 1 and 2?

What is the final velocity v1fof block 1?

KEY IDEAS

Because we assume that the collisions are elastic, we are to

conserve mechanical energy (thus energy losses to sound,

heating, and oscillations of the blocks are negligible).

Because no external horizontal force acts on the blocks, we

are to conserve linear momentum along the xaxis. For these

(a)

(b)

v1i

v1f

v3f

m1m2m3

Collisions in Two Dimensions

When two bodies collide, the impulse between them determines the directions in

which they then travel. In particular, when the collision is not head-on, the bodies

do not end up traveling along their initial axis. For such two-dimensional

collisions in a closed, isolated system, the total linear momentum must still be

conserved:

. (9-77)

If the collision is also elastic (a special case), then the total kinetic energy is also

conserved:

K1iK2iK1fK2f. (9-78)

Equation 9-77 is often more useful for analyzing a two-dimensional collision

if we write it in terms of components on an xy coordinate system. For example,

Fig.9-21 shows a glancing collision (it is not head-on) between a projectile body and a

target body initially at rest.The impulses between the bodies have sent the bodies off

at angles u1and u2to the xaxis,along which the projectile initially traveled. In this situ-

1iP

2iP

1fP

240 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Additional examples, video, and practice available at WileyPLUS

which leads to

(Answer)

Finally, applying Eq. 9-67 to the first collision with this result

and the given v1i, we write

(Answer)

3m2m2

3m2m2

(10 m/s) 5.0 m/s.

v1fm1m2

m1m2

v1i,

m11

3m21

3(6.0 kg) 2.0 kg.

Next, let’s reconsider the first collision, but we have to

be careful with the notation for block 2: its velocity v2fjust

after the first collision is the same as its velocity v2i(5.0 m/s)

just before the second collision. Applying Eq. 9-68 to the

first collision and using the given v1i10 m/s, we have

5.0 m/s 2m1

m1m2

(10 m/s),

v2f2m1

m1m2

v1i,

9-8 COLLISIONS IN TWO DIMENSIONS

After reading this module, you should be able to . . .

9.34 For an isolated system in which a two-dimensional colli-

sion occurs, apply the conservation of momentum along

each axis of a coordinate system to relate the momentum

components along an axis before the collision to the momen-

tum components along the same axis after the collision.

9.35 For an isolated system in which a two-dimensional elastic

collision occurs, (a) apply the conservation of momentum

along each axis of a coordinate system to relate the momen-

tum components along an axis before the collision to the

momentum components along the same axis after the colli-

sion and (b) apply the conservation of total kinetic energy to

relate the kinetic energies before and after the collision.

●If two bodies collide and their motion is not along a single axis

(the collision is not head-on), the collision is two-dimensional.

If the two-body system is closed and isolated, the law of con-

servation of momentum applies to the collision and can be

written as

1iP

2iP

1fP

In component form, the law gives two equations that de-

scribe the collision (one equation for each of the two dimen-

sions). If the collision is also elastic (a special case), the

conservation of kinetic energy during the collision gives a

third equation:

K1iK2iK1fK2f.

Learning Objectives

Key Idea

Figure 9-21 An elastic collision between two

bodies in which the collision is not head-

on. The body with mass m2(the target) is

initially at rest.

1 v1i

v2f

v1f

A glancing collision

that conserves

both momentum and

kinetic energy.

ation we would rewrite Eq.9-77 for components along the xaxis as

m1v1im1v1fcos u1m2v2fcos u2, (9-79)

and along the yaxis as

(9-80)

We can also write Eq. 9-78 (for the special case of an elastic collision) in terms of

speeds:

(kinetic energy). (9-81)

Equations 9-79 to 9-81 contain seven variables: two masses, m1and m2; three

speeds, v1i,v1f, and v2f; and two angles, u1and u2. If we know any four of these

quantities, we can solve the three equations for the remaining three quantities.

2m1v1i

21

2m1v1f

21

2m2v2f

0m1v1fsin u1m2v2fsin u2.

241

9-9 SYSTEMS WITH VARYING MASS: A ROCKET

Checkpoint 9

In Fig. 9-21, suppose that the projectile has an initial momentum of 6 kgm/s, a final

xcomponent of momentum of 4 kgm/s, and a final ycomponent of momentum of

3kgm/s. For the target, what then are (a) the final xcomponent of momentum

and (b) the final ycomponent of momentum?

9-9 SYSTEMS WITH VARYING MASS: A ROCKET

After reading this module, you should be able to . . .

9.36 Apply the first rocket equation to relate the rate at which

the rocket loses mass, the speed of the exhaust products rel-

ative to the rocket, the mass of the rocket, and the accelera-

tion of the rocket.

9.37 Apply the second rocket equation to relate the change in

the rocket’s speed to the relative speed of the exhaust

products and the initial and final mass of the rocket.

9.38 For a moving system undergoing a change in mass at a

given rate, relate that rate to the change in momentum.

●In the absence of external forces a rocket accelerates at an

instantaneous rate given by

Rvrel Ma (first rocket equation),

in which Mis the rocket’s instantaneous mass (including

unexpended fuel), Ris the fuel consumption rate, and vrel is

the fuel’s exhaust speed relative to the rocket. The term Rvrel

is the thrust of the rocket engine.

●For a rocket with constant Rand vrel, whose speed

changes from vito vfwhen its mass changes from Mito Mf,

(second rocket equation).vfvivrel ln Mi

Learning Objectives

Key Ideas

Systems with Varying Mass: A Rocket

So far, we have assumed that the total mass of the system remains constant.

Sometimes, as in a rocket, it does not. Most of the mass of a rocket on its launch-

ing pad is fuel, all of which will eventually be burned and ejected from the nozzle

of the rocket engine. We handle the variation of the mass of the rocket as the

rocket accelerates by applying Newton’s second law, not to the rocket alone but

to the rocket and its ejected combustion products taken together.The mass of this

system does not change as the rocket accelerates.

Finding the Acceleration

Assume that we are at rest relative to an inertial reference frame, watching a

rocket accelerate through deep space with no gravitational or atmospheric drag

242 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Figure 9-22 (a) An accelerating rocket of

mass Mat time t, as seen from an inertial

reference frame. (b) The same but at time

tdt. The exhaust products released dur-

ing interval dt are shown.

forces acting on it. For this one-dimensional motion, let Mbe the mass of the

rocket and vits velocity at an arbitrary time t(see Fig.9-22a).

Figure 9-22bshows how things stand a time interval dt later.The rocket now

has velocity vdv and mass MdM, where the change in mass dM is a negative

quantity. The exhaust products released by the rocket during interval dt have

mass dM and velocity Urelative to our inertial reference frame.

Conserve Momentum. Our system consists of the rocket and the exhaust

products released during interval dt.The system is closed and isolated, so the lin-

ear momentum of the system must be conserved during dt; that is,

PiPf, (9-82)

where the subscripts iand findicate the values at the beginning and end of time

interval dt. We can rewrite Eq. 9-82 as

Mv dM U (MdM)(vdv), (9-83)

where the first term on the right is the linear momentum of the exhaust products

released during interval dt and the second term is the linear momentum of the

rocket at the end of interval dt.

Use Relative Speed. We can simplify Eq. 9-83 by using the relative speed vrel be-

tween the rocket and the exhaust products, which is related to the velocities relative to

the frame with

In symbols, this means

(vdv)vrel U,

or Uvdv vrel. (9-84)

Substituting this result for Uinto Eq. 9-83 yields,with a little algebra,

dM vrel M dv. (9-85)

Dividing each side by dt gives us

(9-86)

We replace dM/dt (the rate at which the rocket loses mass) by R, where Ris the

(positive) mass rate of fuel consumption, and we recognize that dv/dt is the accel-

eration of the rocket.With these changes, Eq. 9-86 becomes

Rvrel Ma (first rocket equation). (9-87)

Equation 9-87 holds for the values at any given instant.

Note the left side of Eq. 9-87 has the dimensions of force (kg/sm/s 

kgm/s2N) and depends only on design characteristics of the rocket engine—

namely, the rate Rat which it consumes fuel mass and the speed vrel with which that

mass is ejected relative to the rocket.We call this term Rvrel the thrust of the rocket

engine and represent it with T.Newton’s second law emerges if we write Eq. 9-87 as

TMa, in which ais the acceleration of the rocket at the time that its mass is M.

Finding the Velocity

How will the velocity of a rocket change as it consumes its fuel? From Eq. 9-85

we have

dv vrel

dM

dt vrel Mdv

dt .



velocity of rocket

relative to frame







velocity of rocket

relative to products







velocity of products

relative to frame



System boundary

(a)

v+ dv

M+dM

System boundary

(b)

–dM

The ejection of mass from

the rocket's rear increases

the rocket's speed.

Integrating leads to

in which Miis the initial mass of the rocket and Mfits final mass. Evaluating the

integrals then gives

(second rocket equation) (9-88)

for the increase in the speed of the rocket during the change in mass from Mito

Mf. (The symbol “ln” in Eq. 9-88 means the natural logarithm.) We see here the

advantage of multistage rockets, in which Mfis reduced by discarding successive

stages when their fuel is depleted. An ideal rocket would reach its destination

with only its payload remaining.

vfvivrel ln Mi

vf

dv vrel Mf

243

REVIEW & SUMMARY

rocket’s mass. However, Mdecreases and aincreases as fuel

is consumed. Because we want the initial value of ahere, we

must use the intial value Miof the mass.

Calculation: We find

(Answer)

To be launched from Earth’s surface, a rocket must have

an initial acceleration greater than . That is, it

must be greater than the gravitational acceleration at the

surface. Put another way, the thrust Tof the rocket engine

must exceed the initial gravitational force on the rocket,

which here has the magnitude Mig, which gives us

(850 kg)(9.8 m/s2)8330 N.

Because the acceleration or thrust requirement is not met

(here T6400 N), our rocket could not be launched from

Earth’s surface by itself; it would require another, more

powerful, rocket.

g9.8 m/s2

aT

6440 N

850 kg 7.6 m/s2.

Sample Problem 9.09 Rocket engine, thrust, acceleration

In all previous examples in this chapter, the mass of a system

is constant (fixed as a certain number). Here is an example of

a system (a rocket) that is losing mass. A rocket whose initial

mass Miis 850 kg consumes fuel at the rate The

speed vrel of the exhaust gases relative to the rocket engine is

2800 m/s.What thrust does the rocket engine provide?

KEY IDEA

Thrust Tis equal to the product of the fuel consumption

rate Rand the relative speed vrel at which exhaust gases are

expelled, as given by Eq. 9-87.

Calculation: Here we find

(Answer)

(b) What is the initial acceleration of the rocket?

KEY IDEA

We can relate the thrust Tof a rocket to the magnitude aof

the resulting acceleration with , where Mis the

TMa

6440 N 6400 N.

TRvrel (2.3 kg/s)(2800 m/s)

R2.3 kg/s.

Additional examples, video, and practice available at WileyPLUS

Center of Mass The center of mass of a system of nparticles is

defined to be the point whose coordinates are given by

(9-5)

or (9-8)

where Mis the total mass of the system.

com 1



i1

mir

xcom 1



i1

mixi, ycom 1



i1

miyi, zcom 1



i1

mizi,

Review & Summary

Newton’s Second Law for a System of Particles The

motion of the center of mass of any system of particles is governed

by Newton’s second law for a system of particles, which is

. (9-14)

Here is the net force of all the external forces acting on the sys-F

net

net Ma

com

tem, Mis the total mass of the system, and is the acceleration

of the system’s center of mass.

:com

244 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

must be conserved (it is a constant), which we can write in vector

form as

, (9-50)

where subscripts iand frefer to values just before and just after the

collision, respectively.

If the motion of the bodies is along a single axis,the collision is

one-dimensional and we can write Eq. 9-50 in terms of velocity

components along that axis:

m1v1im2v2im1v1fm2v2f. (9-51)

If the bodies stick together, the collision is a completely

inelastic collision and the bodies have the same final velocity V

(because they are stuck together).

Motion of the Center of Mass The center of mass of a

closed, isolated system of two colliding bodies is not affected by a

collision. In particular, the velocity of the center of mass can-

not be changed by the collision.

Elastic Collisions in One Dimension An elastic collision

is a special type of collision in which the kinetic energy of a system

of colliding bodies is conserved. If the system is closed and

isolated, its linear momentum is also conserved. For a one-

dimensional collision in which body 2 is a target and body 1 is an

incoming projectile, conservation of kinetic energy and linear

momentum yield the following expressions for the velocities

immediately after the collision:

(9-67)

and (9-68)

Collisions in Two Dimensions If two bodies collide and

their motion is not along a single axis (the collision is not head-on),

the collision is two-dimensional. If the two-body system is closed

and isolated, the law of conservation of momentum applies to the

collision and can be written as

. (9-77)

In component form, the law gives two equations that describe the

collision (one equation for each of the two dimensions). If the col-

lision is also elastic (a special case), the conservation of kinetic en-

ergy during the collision gives a third equation:

K1iK2iK1fK2f. (9-78)

Variable-Mass Systems In the absence of external forces a

rocket accelerates at an instantaneous rate given by

Rvrel Ma (first rocket equation), (9-87)

in which Mis the rocket’s instantaneous mass (including

unexpended fuel), Ris the fuel consumption rate,and vrel is the fuel’s

exhaust speed relative to the rocket. The term Rvrel is the thrust of

the rocket engine. For a rocket with constant Rand vrel,whose speed

changes from vito vfwhen its mass changes from Mito Mf,

(second rocket equation). (9-88)vfvivrel ln Mi

1iP

2iP

1fP

v2f2m1

m1m2

v1i.

v1fm1m2

m1m2

v1i

:com

:1ip

:2ip

:1fp

:2f

tum, and is the impulse due to the force exerted on the body

by the other body in the collision:

(9-30)

If Favg is the average magnitude of during the collision and t

is the duration of the collision, then for one-dimensional motion

JFavg t. (9-35)

When a steady stream of bodies, each with mass mand speed v, col-

lides with a body whose position is fixed, the average force on the

fixed body is

(9-37)

where n/tis the rate at which the bodies collide with the fixed

body, and vis the change in velocity of each colliding body. This

average force can also be written as

(9-40)

where m/tis the rate at which mass collides with the fixed body. In

Eqs. 9-37 and 9-40, vvif the bodies stop upon impact and v

2vif they bounce directly backward with no change in their speed.

Conservation of Linear Momentum If a system is isolated

so that no net external force acts on it, the linear momentum of

the system remains constant:

(closed, isolated system). (9-42)

This can also be written as

(closed, isolated system), (9-43)

where the subscripts refer to the values of at some initial time and

at a later time. Equations 9-42 and 9-43 are equivalent statements of

the law of conservation of linear momentum.

Inelastic Collision in One Dimension In an inelastic

collision of two bodies, the kinetic energy of the two-body

system is not conserved (it is not a constant). If the system is

closed and isolated, the total linear momentum of the system

iP

:constant

Favg m

tv,

Favg  n

tp n

tmv,

(t)

:tf

(t)dt.

(t)J

Linear Momentum and Newton’s Second Law For a sin-

gle particle, we define a quantity called its linear momentum as

, (9-22)

and can write Newton’s second law in terms of this momentum:

(9-23)

For a system of particles these relations become

and (9-25, 9-27)

Collision and Impulse Applying Newton’s second law in

momentum form to a particle-like body involved in a collision

leads to the impulse – linear momentum theorem:

, (9-31, 9-32)

where is the change in the body’s linear momen-p

fp

ip

fp

ip

:J

net dP

dt .P

:Mv

com

net dp

dt .

:mv

245

QUESTIONS

Questions

1Figure 9-23 shows an overhead

view of three particles on which ex-

ternal forces act.The magnitudes and

directions of the forces on two of the

particles are indicated. What are the

magnitude and direction of the force

acting on the third particle if the cen-

ter of mass of the three-particle sys-

tem is (a) stationary, (b) moving at a

constant velocity rightward, and (c) accelerating rightward?

2Figure 9-24 shows an over-

head view of four particles of

equal mass sliding over a fric-

tionless surface at constant

velocity. The directions of the

velocities are indicated; their

magnitudes are equal. Consider

pairing the particles. Which

pairs form a system with a cen-

ter of mass that (a) is stationary,

(b) is stationary and at the ori-

gin,and (c) passes through the origin?

3Consider a box that explodes into two pieces while moving with

a constant positive velocity along an xaxis. If one piece, with mass

m1, ends up with positive velocity , then the second piece, with

mass m2, could end up with (a) a positive velocity (Fig. 9-25a), (b)

a negative velocity (Fig. 9-25b), or (c) zero velocity (Fig. 9-25c).

Rank those three possible results for the second piece according to

the corresponding magnitude of , greatest first.v1

boxes move over a frictionless confectioner’s counter. For each box,

is its linear momentum conserved along the xaxis and the yaxis?

6Figure 9-28 shows four groups of three or four identical particles

that move parallel to either the xaxis or the yaxis,at identical speeds.

Rank the groups according to center-of-mass speed,greatest first.

5 N

3 N

Figure 9-23 Question 1.

c d

y(m)

–2 2 4 –4

–2

x(m)

Figure 9-24 Question 2.

v2v2

v1v1

(b) (c)(a)

Figure 9-25 Question 3.

2F0

4F0

6t0

F F

t t

(a) (b) (c)

3t012t0

2F0

Figure 9-26 Question 4.

xxx

y y y

60°

60° 8 Ν 6 Ν

5 Ν

4 Ν

6 Ν

8 Ν

3 Ν

4 Ν

2 Ν

6 Ν 5 Ν

2 Ν

3 Ν

(a) (b) (c)

Figure 9-27 Question 5.

4Figure 9-26 shows graphs of force magnitude versus time for a

body involved in a collision. Rank the graphs according to the

magnitude of the impulse on the body, greatest first.

5The free-body diagrams in Fig. 9-27 give, from overhead views,

the horizontal forces acting on three boxes of chocolates as the

(a)

(c)

(b)

(d)

Figure 9-28 Question 6.

7A block slides along a frictionless floor and into a stationary sec-

ond block with the same mass. Figure 9-29 shows four choices for a

graph of the kinetic energies Kof the blocks. (a) Determine which

represent physically impossible situations. Of the others, which best

represents (b) an elastic collision and (c) an inelastic collision?

(a)

(b)

(c)

(d)

Figure 9-29 Question 7.

8Figure 9-30 shows a snapshot of

block 1 as it slides along an xaxis on a

frictionless floor, before it undergoes

an elastic collision with stationary

block 2.The figure also shows three possible positions of the center of

mass (com) of the two-block system at the time of the snapshot. (Point

Bis halfway between the centers of the two blocks.) Is block 1 station-

ary,moving forward, or moving backward after the collision if the com

is located in the snapshot at (a) A,(b) B,and (c) C?

1 2

A B C

Figure 9-30 Question 8.

9Two bodies have undergone an

elastic one-dimensional collision

along an xaxis. Figure 9-31 is a graph

of position versus time for those

bodies and for their center of mass.

(a) Were both bodies initially moving,

or was one initially stationary? Which

line segment corresponds to the mo-

tion of the center of mass (b) before the collision and (c) after the col-

lision? (d) Is the mass of the body that was moving faster before the

collision greater than,less than, or equal to that of the other body?

10 Figure 9-32: A block on a horizontal floor is initially either

stationary, sliding in the positive direction of an xaxis, or sliding in

246 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

the negative direction of that axis. Then the block explodes into

two pieces that slide along the xaxis. Assume the block and the

two pieces form a closed, isolated system. Six choices for a graph of

the momenta of the block and the pieces are given, all versus time

t. Determine which choices represent physically impossible situa-

tions and explain why.

11 Block 1 with mass m1slides

along an xaxis across a frictionless

floor and then undergoes an elastic

collision with a stationary block 2 with

mass m2. Figure 9-33 shows a plot of

position xversus time tof block 1 until

the collision occurs at position xcand

time tc.In which of the lettered regions

on the graph will the plot be contin-

ued (after the collision) if (a) m1m2

and (b) m1m2? (c) Along which of

the numbered dashed lines will the

plot be continued if m1m2?

12 Figure 9-34 shows four graphs of

position versus time for two bodies

and their center of mass. The two

bodies form a closed, isolated system

and undergo a completely inelastic,

one-dimensional collision on an xaxis.

In graph 1, are (a) the two bodies and

(b) the center of mass moving in the

positive or negative direction of the x

axis? (c) Which of the graphs corre-

spond to a physically impossible situ-

ation? Explain.

2 3

Figure 9-31 Question 9.

(a)

(d)

(b)

(e)

(c)

(f)

Figure 9-32 Question 10.

Figure 9-33 Question 11.

x x

t t

(1) (2)

(3) (4)

Figure 9-34 Question 12.

Module 9-1 Center of Mass

•1 A 2.00 kg particle has the xy coordinates (1.20 m, 0.500 m),

and a 4.00 kg particle has the xy coordinates (0.600 m, 0.750 m).

Both lie on a horizontal plane. At what (a) xand (b) ycoordinates

must you place a 3.00 kg particle such that the center of mass of the

three-particle system has the coor-

dinates (0.500 m, 0.700 m)?

•2 Figure 9-35 shows a three-par-

ticle system, with masses m13.0

kg, m24.0 kg, and m3 8.0 kg.

The scales on the axes are set by

xs2.0 m and ys2.0 m.What are

(a) the xcoordinate and (b) the y

coordinate of the system’s center

of mass? (c) If m3is gradually in-

creased, does the center of mass of the system shift toward or away

from that particle, or does it remain stationary?

••3 Figure 9-36 shows a slab with dimensions d111.0 cm, d2

2.80 cm, and d313.0 cm. Half the slab consists of aluminum (den-

Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

•–••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday

Problems

y(m)

x(m)

0xs

Figure 9-35 Problem 2.

Aluminum

Iron Midpoint

2d1

d1d1

sity 2.70 g/cm3) and half consists of iron (density 7.85 g/cm3).

What are (a) the xcoordinate, (b) the ycoordinate, and (c) the zco-

ordinate of the slab’s center of mass?

Figure 9-36 Problem 3.

247

PROBLEMS

••4 In Fig. 9-37, three uniform thin rods,

each of length L22 cm, form an in-

verted U. The vertical rods each have a

mass of 14 g; the horizontal rod has a

mass of 42 g. What are (a) the xcoordi-

nate and (b) the ycoordinate of the sys-

tem’s center of mass?

••5 What are (a) the xcoordinate and

(b) the ycoordinate of the center of mass

for the uniform plate shown in Fig. 9-38 if

L5.0 cm?

Module 9-2 Newton’s Second Law for a System of Particles

•9 A stone is dropped at t0.A second stone, with twice the

mass of the first, is dropped from the same point at

t100 ms. (a) How far below the release point is the center of

mass of the two stones at t300 ms? (Neither stone has yet

reached the ground.) (b) How fast is the center of mass of the two-

stone system moving at that time?

•10 A 1000 kg automobile is at rest at a traffic signal. At the in-

stant the light turns green, the automobile starts to move with a

constant acceleration of 4.0 m/s2. At the same instant a 2000 kg

truck, traveling at a constant speed of 8.0 m/s, overtakes and passes

the automobile. (a) How far is the com of the automobile–truck

system from the traffic light at t3.0 s? (b) What is the speed of

the com then?

•11 A big olive (m0.50 kg) lies at the origin of an xy

coordinate system, and a big Brazil nut (M1.5 kg) lies at the

point (1.0, 2.0) m. At t0, a force begins to

act on the olive, and a force begins to act on

the nut. In unit-vector notation, what is the displacement of the

center of mass of the olive–nut system at t4.0 s, with respect to

its position at t0?

•12 Two skaters, one with mass 65 kg and the other with mass

40 kg, stand on an ice rink holding a pole of length 10 m and neg-

ligible mass. Starting from the ends of the pole, the skaters pull

themselves along the pole until they meet. How far does the 40

kg skater move?

••13 A shell is shot with an initial velocity of 20 m/s, at

an angle of with the horizontal. At the top of the trajec-

tory, the shell explodes into two fragments of equal mass (Fig.

9-42). One fragment, whose speed immediately after the explo-

sion is zero, falls vertically. How far from the gun does the other

fragment land, assuming that the terrain is level and that air drag

is negligible?



060

SSM

n(3.0i

ˆ2.0j

ˆ) N

o(2.0i

ˆ3.0j

ˆ) N

ILW

Figure 9-37 Problem 4.

Figure 9-38 Problem 5.

••6 Figure 9-39 shows a cubical box that

has been constructed from uniform metal

plate of negligible thickness. The box is

open at the top and has edge length L

40 cm. Find (a) the xcoordinate, (b) the y

coordinate, and (c) the zcoordinate of

the center of mass of the box.

•••7 In the ammonia (NH3) mole-

cule of Fig. 9-40, three hydrogen (H)

atoms form an equilateral triangle, with

the center of the triangle at distance d

9.40 1011 m from each hydrogen

atom. The nitrogen (N) atom is at the

apex of a pyramid, with the three hydro-

gen atoms forming the base. The nitro-

gen-to-hydrogen atomic mass ratio is

13.9, and the nitrogen-to-hydrogen dis-

tance is L10.14 1011 m. What are

the (a) xand (b) ycoordinates of the

molecule’s center of mass?

•••8 A uniform soda can of mass

0.140 kg is 12.0 cm tall and filled with

0.354 kg of soda (Fig. 9-41). Then small

holes are drilled in the top and bottom

(with negligible loss of metal) to drain

the soda. What is the height hof the

com of the can and contents (a) initially

and (b) after the can loses all the soda?

drains out? (d) If xis the height of the

remaining soda at any given instant,

find xwhen the com reaches its lowest

point.

ILW

LOy

Figure 9-39 Problem 6.

Figure 9-40 Problem 7.

Figure 9-41 Problem 8.

Explosion

Figure 9-42 Problem 13.

••14 In Figure 9-43, two particles are launched from the origin of

the coordinate system at time t0. Particle 1 of mass m15.00 g is

shot directly along the xaxis on a frictionless floor, with constant

speed 10.0 m/s. Particle 2 of mass m23.00 g is shot with a velocity

of magnitude 20.0 m/s, at an upward angle such that it always stays

directly above particle 1. (a) What is the maximum height Hmax

reached by the com of the two-particle system? In unit-vector no-

tation, what are the (b) velocity and (c) acceleration of the com

when the com reaches Hmax?

Figure 9-43 Problem 14.

248 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

••15 Figure 9-44 shows an arrangement with an air track, in which

a cart is connected by a cord to a hanging block. The cart has mass

m10.600 kg, and its center is initially at xy coordinates (0.500

m, 0 m); the block has mass m20.400 kg,and its center is initially at

xy coordinates (0, 0.100 m).The mass of the cord and pulley are neg-

ligible. The cart is released from rest, and both cart and block move

until the cart hits the pulley. The friction between the cart and the air

track and between the pulley and its axle is negligible. (a) In unit-vec-

tor notation, what is the acceleration of the center of mass of the

cart–block system? (b) What is the velocity of the com as a function

of time t? (c) Sketch the path taken by the com. (d) If the path is

curved, determine whether it bulges upward to the right or downward

to the left, and if it is straight, find the angle between it and the xaxis.

••21 A 0.30 kg softball has a velocity of 15 m/s at an angle of 35be-

low the horizontal just before making contact with the bat.What is the

magnitude of the change in momentum of the ball while in contact

with the bat if the ball leaves with a velocity of (a) 20 m/s, vertically

downward,and (b) 20 m/s,horizontally back toward the pitcher?

••22 Figure 9-47 gives an overhead

view of the path taken by a 0.165 kg

cue ball as it bounces from a rail of a

pool table. The ball’s initial speed is

2.00 m/s, and the angle u1is 30.0.The

bounce reverses the ycomponent of

the ball’s velocity but does not alter

the xcomponent.What are (a) angle

u2and (b) the change in the ball’s lin-

ear momentum in unit-vector nota-

tion? (The fact that the ball rolls is ir-

relevant to the problem.)

Module 9-4 Collision and Impulse

•23 Until his seventies, Henri LaMothe (Fig. 9-48) excited

audiences by belly-flopping from a height of 12 m into 30 cm of

water.Assuming that he stops just as he reaches the bottom of the

water and estimating his mass, find the magnitude of the impulse

on him from the water.

Figure 9-44 Problem 15.

•••16 Ricardo, of mass 80 kg, and Carmelita, who is lighter,

are enjoying Lake Merced at dusk in a 30 kg canoe.When the ca-

noe is at rest in the placid water, they exchange seats, which are

3.0 m apart and symmetrically lo-

cated with respect to the canoe’s

center. If the canoe moves 40 cm

horizontally relative to a pier post,

what is Carmelita’s mass?

•••17 In Fig. 9-45a, a 4.5 kg dog

stands on an 18 kg flatboat at dis-

tance D6.1 m from the shore. It

walks 2.4 m along the boat toward

shore and then stops. Assuming no

friction between the boat and the wa-

ter, find how far the dog is then from

the shore.(Hint: See Fig.9-45b.)

Module 9-3 Linear Momentum

•18 A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical

wall and rebounds with speed 2.0 m/s. What is the magnitude of the

change in its linear momentum?

•19 A 2100 kg truck traveling north at 41 km/h turns east

and accelerates to 51 km/h. (a) What is the change in the truck’s

kinetic energy? What are the (b) magnitude and (c) direction of

the change in its momentum?

••20 At time t0, a ball is

struck at ground level and sent over

level ground.The momentum pver-

sus tduring the flight is given by Fig.

9-46 (with and

). At what initial

angle is the ball launched? (Hint:

Find a solution that does not

require you to read the time of the

low point of the plot.)

p14.0 kgm/s

p06.0 kgm/s

ILW

Dog's displacement dd

Boat's displacement db

(b)

(a)

Figure 9-45 Problem 17.

p (kg • m/s)

p1531 4 2

t (s)

Figure 9-46 Problem 20.

Figure 9-47 Problem 22.

George Long/Getty Images, Inc.

Figure 9-48 Problem 23. Belly-flopping into 30 cm of water.

•24 In February 1955, a paratrooper fell 370 m from an air-

plane without being able to open his chute but happened to land in

snow, suffering only minor injuries. Assume that his speed at im-

pact was 56 m/s (terminal speed), that his mass (including gear)

was 85 kg, and that the magnitude of the force on him from the

249

PROBLEMS

12.0 m/s and angle u135.0. Just

after, it is traveling directly upward

with velocity of magnitude 10.0

m/s. The duration of the collision is

2.00 ms. What are the (a) magni-

snow was at the survivable limit of 1.2 105N. What are (a) the

minimum depth of snow that would have stopped him safely and

(b) the magnitude of the impulse on him from the snow?

•25 A 1.2 kg ball drops vertically onto a floor, hitting with a

speed of 25 m/s. It rebounds with an initial speed of 10 m/s. (a)

What impulse acts on the ball during the contact? (b) If the ball is

in contact with the floor for 0.020 s, what is the magnitude of the

average force on the floor from the ball?

•26 In a common but dangerous prank, a chair is pulled away as

a person is moving downward to sit on it, causing the victim to land

hard on the floor. Suppose the victim falls by 0.50 m, the mass that

moves downward is 70 kg,and the collision on the floor lasts 0.082 s.

What are the magnitudes of the (a) impulse and (b) average force

acting on the victim from the floor during the collision?

•27 A force in the negative direction of an xaxis is applied

for 27 ms to a 0.40 kg ball initially moving at 14 m/s in the positive

direction of the axis. The force varies in magnitude, and the im-

pulse has magnitude 32.4 Ns.What are the ball’s (a) speed and (b)

direction of travel just after the force is applied? What are (c) the

average magnitude of the force and (d) the direction of the im-

pulse on the ball?

•28 In tae-kwon-do, a hand is slammed down onto a target

at a speed of 13 m/s and comes to a stop during the 5.0 ms collision.

Assume that during the impact the hand is independent of the arm

and has a mass of 0.70 kg. What are the magnitudes of the (a) im-

pulse and (b) average force on the hand from the target?

•29 Suppose a gangster sprays Superman’s chest with 3 g bullets

at the rate of 100 bullets/min, and the speed of each bullet is 500

m/s. Suppose too that the bullets rebound straight back with no

change in speed. What is the magnitude of the average force on

Superman’s chest?

••30 Two average forces. A steady stream of 0.250 kg snowballs is

shot perpendicularly into a wall at a speed of 4.00 m/s. Each ball

sticks to the wall. Figure 9-49 gives the magnitude Fof the force on

the wall as a function of time tfor two of the snowball impacts.

Impacts occur with a repetition time interval tr50.0 ms, last a du-

ration time interval td10 ms, and produce isosceles triangles on

the graph, with each impact reaching a force maximum Fmax 200 N.

During each impact, what are the magnitudes of (a) the impulse and

(b) the average force on the wall? (c) During a time interval of many

impacts,what is the magnitude of the average force on the wall?

SSM

are the magnitudes of the (c) impulse

and (d) average force (assuming the

same stopping time)?

••32 A 5.0 kg toy car can move

along an xaxis; Fig. 9-50 gives Fxof

the force acting on the car, which be-

gins at rest at time t0. The scale on

the Fxaxis is set by In

unit-vector notation, what is at (a)

t4.0 s and (b) t7.0 s, and (c)

what is at t9.0 s?

••33 Figure 9-51 shows a 0.300

kg baseball just before and just after

it collides with a bat. Just before, the

ball has velocity of magnitudev

v



Fxs 5.0 N.

Δtr

ΔtdΔtd

Fmax

Figure 9-49 Problem 30.

24 86

–

Fxs

t (s)

Fx(N)

Figure 9-50 Problem 32.

v2v1

Figure 9-51 Problem 33.

Stephen Dalton/Photo Researchers, Inc.

Figure 9-52 Problem 34. Lizard running across water.

••31 Jumping up before the elevator hits. After the cable

snaps and the safety system fails, an elevator cab free-falls from a

height of 36 m. During the collision at the bottom of the elevator

shaft, a 90 kg passenger is stopped in 5.0 ms. (Assume that neither the

passenger nor the cab rebounds.) What are the magnitudes of the (a)

impulse and (b) average force on the passenger during the collision?

If the passenger were to jump upward with a speed of 7.0 m/s relative

to the cab floor just before the cab hits the bottom of the shaft, what

tude and (b) direction (relative to the positive direction of the x

axis) of the impulse on the ball from the bat? What are the (c)

magnitude and (d) direction of the average force on the ball from

the bat?

••34 Basilisk lizards can run across the top of a water sur-

face (Fig. 9-52). With each step, a lizard first slaps its foot against

the water and then pushes it down into the water rapidly enough to

form an air cavity around the top of the foot. To avoid having to

pull the foot back up against water drag in order to complete the

step, the lizard withdraws the foot before water can flow into the

air cavity. If the lizard is not to sink, the average upward impulse

on the lizard during this full action of slap, downward push, and

withdrawal must match the downward impulse due to the gravita-

tional force. Suppose the mass of a basilisk lizard is 90.0 g, the mass

of each foot is 3.00 g, the speed of a foot as it slaps the water is

1.50 m/s, and the time for a single step is 0.600 s. (a) What is the

magnitude of the impulse on the lizard during the slap? (Assume

this impulse is directly upward.) (b) During the 0.600 s duration of

a step, what is the downward impulse on the lizard due to the gravi-

tational force? (c) Which action, the slap or the push, provides the

primary support for the lizard, or are they approximately equal in

their support?

250 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

= 0

Figure 9-57 Problem 44.

••35 Figure 9-53 shows an

approximate plot of force mag-

nitude Fversus time tduring the

collision of a 58 g Superball with

a wall. The initial velocity of the

ball is 34 m/s perpendicular to

the wall; the ball rebounds di-

rectly back with approximately

the same speed, also perpendi-

cular to the wall. What is Fmax,

the maximum magnitude of the

force on the ball from the wall during the collision?

••36 A 0.25 kg puck is initially stationary on an ice surface with

negligible friction. At time t0, a horizontal force begins to

move the puck.The force is given by (12.0 , with

in newtons and tin seconds, and it acts until its magnitude is

zero. (a) What is the magnitude of the impulse on the puck from

the force between t0.500 s and t1.25 s? (b) What is the

change in momentum of the puck between t0 and the instant

at which F0?

••37 A soccer player kicks a soccer ball of mass 0.45 kg that

is initially at rest. The foot of the player is in contact with the ball

for 3.0 103s,and the force of the kick is given by

F(t)[(6.0 106)t(2.0 109)t2] N

for 0 t3.0 103s, where tis in seconds. Find the magnitudes

of (a) the impulse on the ball due to the kick, (b) the average force

on the ball from the player’s foot during the period of contact,

period of contact, and (d) the ball’s velocity immediately after it

loses contact with the player’s foot.

••38 In the overhead view of Fig.

9-54, a 300 g ball with a speed vof

6.0 m/s strikes a wall at an angle u

of 30and then rebounds with the

same speed and angle. It is in con-

tact with the wall for 10 ms. In unit-

vector notation, what are (a) the

impulse on the ball from the wall

and (b) the average force on the wall from the ball?

SSM

3.00t2)i

locity that the explosion gives the rest of the rocket. (2) Next, at

time t0.80 s, block Ris shot to the right with a speed of 3.00 m/s

relative to the velocity that block Cthen has. At t2.80 s, what

are (a) the velocity of block Cand (b) the position of its center?

••42 An object, with mass mand speed vrelative to an observer,

explodes into two pieces, one three times as massive as the other;

the explosion takes place in deep space. The less massive piece

stops relative to the observer. How much kinetic energy is added

to the system during the explosion, as measured in the observer’s

reference frame?

••43 In the Olympiad of 708 B.C., some athletes competing in

the standing long jump used handheld weights called halteres to

lengthen their jumps (Fig.9-56).The weights were swung up in front

just before liftoff and then swung down and thrown backward dur-

ing the flight. Suppose a modern 78 kg long jumper similarly uses

two 5.50 kg halteres, throwing them horizontally to the rear at his

maximum height such that their horizontal velocity is zero rela-

tive to the ground. Let his liftoff velocity be m/s

with or without the halteres, and assume that he lands at the liftoff

level. What distance would the use of the halteres add to his range?

: (9.5i

ˆ4.0j

ˆ)

Module 9-5 Conservation of Linear Momentum

•39 A 91 kg man lying on a surface of negligible friction

shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s.

What speed does the man acquire as a result?

•40 A space vehicle is traveling at 4300 km/h relative to Earth

when the exhausted rocket motor (mass 4m) is disengaged and

sent backward with a speed of 82 km/h relative to the command

module (mass m).What is the speed of the command module rel-

ative to Earth just after the separation?

••41 Figure 9-55 shows a two-ended “rocket” that is initially sta-

tionary on a frictionless floor, with its center at the origin of an x

axis.The rocket consists of a central block C(of mass M6.00 kg)

and blocks Land R(each of mass m2.00 kg) on the left and

right sides. Small explosions can

shoot either of the side blocks away

from block Cand along the xaxis.

Here is the sequence: (1) At time t

0, block Lis shot to the left with a

speed of 3.00 m/s relative to the ve-

SSM

θ θ

Figure 9-54 Problem 38.

L R

Figure 9-55 Problem 41.

••44 In Fig. 9-57, a stationary block explodes into two pieces L

and Rthat slide across a frictionless floor and then into regions with

friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a

coefficient of kinetic friction mL0.40 and slides to a stop in distance

dL0.15 m. Piece Rencounters a coefficient of kinetic friction mR

0.50 and slides to a stop in distance dR0.25 m. What was the mass

of the block?

Réunion des Musées Nationaux/

Art Resource

Figure 9-56 Problem 43.

Fmax

t(ms)

F (N)

Figure 9-53 Problem 35.

••45 A 20.0 kg body is moving through space in the

positive direction of an xaxis with a speed of 200 m/s when, due

to an internal explosion, it breaks into three parts. One part, with a

mass of 10.0 kg, moves away from the point of explosion with

a speed of 100 m/s in the positive ydirection.A second part, with a

mass of 4.00 kg, moves in the negative xdirection with a speed of

500 m/s. (a) In unit-vector notation, what is the velocity of the third

part? (b) How much energy is released in the explosion? Ignore ef-

fects due to the gravitational force.

••46 A 4.0 kg mess kit sliding on a frictionless surface explodes

into two 2.0 kg parts: 3.0 m/s, due north, and 5.0 m/s, 30north of

east.What is the original speed of the mess kit?

WWWSSM

251

PROBLEMS

••47 A vessel at rest at the origin of an xy coordinate system ex-

plodes into three pieces. Just after the explosion, one piece, of mass

m, moves with velocity (30 m/s) and a second piece, also of mass

m, moves with velocity (30 m/s) . The third piece has mass 3m.

Just after the explosion, what are the (a) magnitude and (b) direc-

tion of the velocity of the third piece?

•••48 Particle Aand particle Bare held together with a com-

pressed spring between them. When they are released, the spring

pushes them apart, and they then fly off in opposite directions, free of

the spring.The mass of Ais 2.00 times the mass of B, and the energy

stored in the spring was 60 J.Assume that the spring has negligible

mass and that all its stored energy is transferred to the particles.

Once that transfer is complete, what are the kinetic energies of (a)

particle Aand (b) particle B?

Module 9-6 Momentum and Kinetic Energy in Collisions

•49 A bullet of mass 10 g strikes a ballistic pendulum of mass

2.0 kg.The center of mass of the pendulum rises a vertical distance

of 12 cm. Assuming that the bullet remains embedded in the pen-

dulum, calculate the bullet’s initial speed.

•50 A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden

block at rest on a frictionless surface.The bullet emerges, traveling

in the same direction with its speed reduced to 428 m/s. (a) What is

the resulting speed of the block? (b) What is the speed of the

bullet–block center of mass?

••51 In Fig. 9-58a, a 3.50 g bullet is fired horizontally at two

blocks at rest on a frictionless table.The bullet passes through block

1 (mass 1.20 kg) and embeds itself in block 2 (mass 1.80 kg). The

blocks end up with speeds v10.630 m/s and v21.40 m/s (Fig.

9-58b). Neglecting the material removed from block 1 by the bullet,

find the speed of the bullet as it (a) leaves and (b) enters block 1.

collisions (CVC). (b) What percent of the original kinetic energy is

lost if the car hits a 300 kg camel? (c) Generally, does the percent

loss increase or decrease if the animal mass decreases?

••54 A completely inelastic collision occurs between two balls of

wet putty that move directly toward each other along a vertical

axis. Just before the collision, one ball, of mass 3.0 kg, is moving up-

ward at 20 m/s and the other ball, of mass 2.0 kg, is moving down-

ward at 12 m/s. How high do the combined two balls of putty rise

above the collision point? (Neglect air drag.)

••55 A 5.0 kg block with a speed of 3.0 m/s collides with a 10

kg block that has a speed of 2.0 m/s in the same direction.After the

collision, the 10 kg block travels in the original direction with a

speed of 2.5 m/s. (a) What is the velocity of the 5.0 kg block imme-

diately after the collision? (b) By how much does the total kinetic

energy of the system of two blocks change because of the colli-

sion? (c) Suppose, instead, that the 10 kg block ends up with a

speed of 4.0 m/s. What then is the change in the total kinetic en-

ergy? (d) Account for the result you obtained in (c).

••56 In the “before” part of Fig. 9-60, car A(mass 1100 kg) is

stopped at a traffic light when it is rear-ended by car B(mass

1400 kg). Both cars then slide with locked wheels until the fric-

tional force from the slick road (with a low mkof 0.13) stops them,

at distances dA8.2 m and dB6.1 m.What are the speeds of (a)

car Aand (b) car Bat the start of the sliding, just after the colli-

sion? (c) Assuming that linear momentum is conserved during

the collision, find the speed of car Bjust before the collision.

(d) Explain why this assumption may be invalid.

ILW

••52 In Fig. 9-59, a 10 g bullet

moving directly upward at 1000 m/s

strikes and passes through the cen-

ter of mass of a 5.0 kg block initially

at rest. The bullet emerges from the

block moving directly upward at 400

m/s. To what maximum height does

the block then rise above its initial

position?

••53 In Anchorage, collisions of a vehicle with a moose are so

common that they are referred to with the abbreviation MVC.

Suppose a 1000 kg car slides into a stationary 500 kg moose on a

very slippery road, with the moose being thrown through the wind-

shield (a common MVC result). (a) What percent of the original

kinetic energy is lost in the collision to other forms of energy? A

similar danger occurs in Saudi Arabia because of camel–vehicle

1 2

Frictionless

(a)

(b)

v1v2

Figure 9-58 Problem 51.

Bullet

Figure 9-59 Problem 52.

••57 In Fig. 9-61, a ball of mass

Figure 9-60 Problem 56.

Before

After

vim

Figure 9-61 Problem 57.

m60 g is shot with speed vi22

m/s into the barrel of a spring gun of

mass M240 g initially at rest on a

frictionless surface.The ball sticks in



the barrel at the point of maximum compression of the spring.

Assume that the increase in thermal energy due to friction be-

tween the ball and the barrel is negligible. (a) What is the speed of

the spring gun after the ball stops in the barrel? (b) What fraction

of the initial kinetic energy of the ball is stored in the spring?

•••58 In Fig. 9-62, block 2 (mass 1.0

kg) is at rest on a frictionless surface

and touching the end of an un-

stretched spring of spring constant

200 N/m.The other end of the spring

is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v14.0

m/s,collides with block 2, and the two blocks stick together.When the

blocks momentarily stop, by what distance is the spring compressed?

1 2

Figure 9-62 Problem 58.

252 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Frictionless

Figure 9-67 Problem 68.

Basketball

Baseball

(a) Before (b) After

Figure 9-68 Problem 69.

1 2

2d d

Figure 9-69 Problem 70.

•••59 In Fig. 9-63, block 1 (mass 2.0 kg) is moving rightward at

10 m/s and block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s.

The surface is frictionless, and a spring with a spring constant of

1120 N/m is fixed to block 2.When the blocks collide, the compres-

sion of the spring is maximum at the instant the blocks have the

same velocity. Find the maximum compression.

ILW friction is 0.50; there they stop. How far into that region do (a)

block 1 and (b) block 2 slide?

••67 In Fig. 9-66, particle 1 of mass

m10.30 kg slides rightward along

an xaxis on a frictionless floor with a

speed of 2.0 m/s.When it reaches x

0, it undergoes a one-dimensional

elastic collision with stationary parti-

cle 2 of mass m20.40 kg.When par-

ticle 2 then reaches a wall at xw70 cm, it bounces from the wall

with no loss of speed. At what position on the xaxis does particle 2

then collide with particle 1?

••68 In Fig. 9-67, block 1 of mass m1slides from rest along a

frictionless ramp from height h2.50 m and then collides with

stationary block 2, which has mass m22.00m1.After the collision,

block 2 slides into a region where the coefficient of kinetic friction

mkis 0.500 and comes to a stop in distance dwithin that region.

What is the value of distance dif the collision is (a) elastic and (b)

completely inelastic?



Module 9-7 Elastic Collisions in One Dimension

•60 In Fig. 9-64, block A(mass 1.6

kg) slides into block B(mass 2.4 kg),

along a frictionless surface. The direc-

tions of three velocities before (i) and

after ( f) the collision are indicated;

the corresponding speeds are vAi

5.5 m/s, vBi 2.5 m/s, and vBf 4.9

m/s. What are the (a) speed and (b)

direction (left or right) of velocity

? (c) Is the collision elastic?

•61 A cart with mass 340 g

moving on a frictionless linear air track at an initial speed of 1.2 m/s

undergoes an elastic collision with an initially stationary cart of un-

known mass.After the collision, the first cart continues in its origi-

nal direction at 0.66 m/s. (a) What is the mass of the second cart?

(b) What is its speed after impact? (c) What is the speed of the two-

cart center of mass?

•62 Two titanium spheres approach each other head-on with the

same speed and collide elastically. After the collision, one of the

spheres, whose mass is 300 g, remains at rest. (a) What is the mass

of the other sphere? (b) What is the speed of the two-sphere center

of mass if the initial speed of each sphere is 2.00 m/s?

••63 Block 1 of mass m1slides along a frictionless floor and into a

one-dimensional elastic collision with stationary block 2 of mass

m23m1. Prior to the collision, the center of mass of the two-

block system had a speed of 3.00 m/s. Afterward, what are the

speeds of (a) the center of mass and (b) block 2?

••64 A steel ball of mass 0.500 kg

is fastened to a cord that is 70.0 cm long

and fixed at the far end.The ball is then

released when the cord is horizontal

(Fig. 9-65). At the bottom of its path,

the ball strikes a 2.50 kg steel block ini-

tially at rest on a frictionless surface.

The collision is elastic. Find (a) the

speed of the ball and (b) the speed of

the block, both just after the collision.

••65 A body of mass 2.0 kg makes an elastic collision with

another body at rest and continues to move in the original

direction but with one-fourth of its original speed. (a) What is the

mass of the other body? (b) What is the speed of the two-body cen-

ter of mass if the initial speed of the 2.0 kg body was 4.0 m/s?

••66 Block 1, with mass m1and speed 4.0 m/s, slides along an x

axis on a frictionless floor and then undergoes a one-dimensional

elastic collision with stationary block 2, with mass m20.40m1.The

two blocks then slide into a region where the coefficient of kinetic

SSM





1 2

Figure 9-63 Problem 59.

vAf = ? vBf

vAi vBi

Figure 9-64 Problem 60.

Figure 9-65 Problem 64.

x (cm)

0xw

1 2

Figure 9-66 Problem 67.

•••69 A small ball of

mass mis aligned above a larger ball

of mass M0.63 kg (with a slight

separation, as with the baseball and

basketball of Fig. 9-68a), and the

two are dropped simultaneously

from a height of h1.8 m.

(Assume the radius of each ball is

negligible relative to h.) (a) If the

larger ball rebounds elastically

from the floor and then the small

ball rebounds elastically from the

larger ball, what value of mresults

in the larger ball stopping when it

collides with the small ball? (b)

What height does the small ball

then reach (Fig. 9-68b)?

•••70 In Fig. 9-69, puck 1 of mass m10.20 kg is sent sliding

across a frictionless lab bench, to undergo a one-dimensional elas-

tic collision with stationary puck 2. Puck 2 then slides off the bench

and lands a distance dfrom the base of the bench. Puck 1 rebounds

from the collision and slides off the opposite edge of the bench,

landing a distance 2dfrom the base of the bench. What is the mass

of puck 2? (Hint: Be careful with signs.)

253

PROBLEMS

Module 9-8 Collisions in Two Dimensions

••71 In Fig. 9-21, projectile particle 1 is an alpha particle and

target particle 2 is an oxygen nucleus. The alpha particle is scattered

at angle u164.0and the oxygen nucleus recoils with speed 1.20 

105m/s and at angle u251.0. In atomic mass units, the mass of the

alpha particle is 4.00 u and the mass of the oxygen nucleus is 16.0 u.

What are the (a) final and (b) initial speeds of the alpha particle?

••72 Ball B, moving in the positive direction of an xaxis at speed

v, collides with stationary ball Aat the origin. Aand Bhave differ-

ent masses.After the collision, Bmoves in the negative direction of

the yaxis at speed v/2. (a) In what direction does Amove?

(b) Show that the speed of Acannot be determined from the given

information.

••73 After a completely inelastic collision, two objects of the same

mass and same initial speed move away together at half their initial

speed. Find the angle between the initial velocities of the objects.

••74 Two 2.0 kg bodies, Aand B, collide. The velocities before the

collision are and m/s.After

the collision, What are (a) the final velocity

of Band (b) the change in the total kinetic energy (including sign)?

••75 A projectile proton with a speed of 500 m/s collides elasti-

cally with a target proton initially at rest. The two protons then

move along perpendicular paths, with the projectile path at 60

from the original direction.After the collision, what are the speeds

of (a) the target proton and (b) the projectile proton?

Module 9-9 Systems with Varying Mass: A Rocket

•76 A 6090 kg space probe moving nose-first toward Jupiter at

105 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg

of exhaust at a speed of 253 m/s relative to the space probe.What is

the final velocity of the probe?

•77 In Fig. 9-70, two long barges are moving in the same

direction in still water, one with a speed of 10 km/h and the other

with a speed of 20 km/h.While they are passing each other, coal is

shoveled from the slower to the faster one at a rate of 1000 kg/min.

How much additional force must be provided by the driving en-

gines of (a) the faster barge and (b) the slower barge if neither is to

change speed? Assume that the shoveling is always perfectly side-

ways and that the frictional forces between the barges and the water

do not depend on the mass of the barges.

SSM

20j

ˆ) m/s.v

:

A(5.0i

ˆ

(10i

ˆ5.0j

ˆ)v

Bv

A(15i

ˆ30j

ˆ) m/s



ILW

certain interval. What must be the rocket’s mass ratio (ratio of ini-

tial to final mass) over that interval if the rocket’s original speed

relative to the inertial frame is to be equal to (a) the exhaust speed

(speed of the exhaust products relative to the rocket) and (b) 2.0

times the exhaust speed?

•79 A rocket that is in deep space and initially at rest

relative to an inertial reference frame has a mass of 2.55 105kg,

of which 1.81 105kg is fuel. The rocket engine is then fired for

250 s while fuel is consumed at the rate of 480 kg/s. The speed of

the exhaust products relative to the rocket is 3.27 km/s. (a) What is

the rocket’s thrust? After the 250 s firing, what are (b) the mass

and (c) the speed of the rocket?

Additional Problems

80 An object is tracked by a radar station and determined to have

a position vector given by (3500 160t)2700 300 , with

in meters and tin seconds.The radar station’s xaxis points east,

its yaxis north, and its zaxis vertically up. If the object is a 250 kg

meteorological missile, what are (a) its linear momentum, (b) its

direction of motion, and (c) the net force on it?

81 The last stage of a rocket, which is traveling at a speed of

7600 m/s, consists of two parts that are clamped together: a rocket

case with a mass of 290.0 kg and a payload capsule with a mass of

150.0 kg. When the clamp is released, a compressed spring causes

the two parts to separate with a relative speed of 910.0 m/s. What

are the speeds of (a) the rocket case and (b) the payload after they

have separated? Assume that all velocities are along the same line.

Find the total kinetic energy of the two parts (c) before and (d) after

they separate. (e) Account for the difference.

82 Pancake collapse of a tall

building. In the section of a tall

building shown in Fig. 9-71a, the in-

frastructure of any given floor K

must support the weight Wof all

higher floors. Normally the infra-

structure is constructed with a

safety factor sso that it can with-

stand an even greater downward

force of sW. If, however, the support

columns between Kand Lsuddenly

collapse and allow the higher floors to free-fall together onto floor

K(Fig. 9-71b), the force in the collision can exceed sW and, after a

brief pause, cause Kto collapse onto floor J, which collapses on

floor I, and so on until the ground is reached. Assume that the

floors are separated by and have the same mass.Also as-

sume that when the floors above Kfree-fall onto K, the collision

lasts 1.5 ms. Under these simplified conditions, what value must the

safety factor sexceed to prevent pancake collapse of the building?

83 “Relative” is an important

word. In Fig. 9-72, block Lof mass

mL1.00 kg and block Rof mass

mR0.500 kg are held in place with

a compressed spring between them.

When the blocks are released, the spring sends them sliding across

a frictionless floor. (The spring has negligible mass and falls to the

floor after the blocks leave it.) (a) If the spring gives block La re-

lease speed of 1.20 m/s relative to the floor, how far does block R

travel in the next 0.800 s? (b) If, instead, the spring gives block La

release speed of 1.20 m/s relative to the velocity that the spring

gives block R, how far does block Rtravel in the next 0.800 s?

d4.0 m



ILWSSM

Figure 9-70 Problem 77.

(a) (b)

Figure 9-71 Problem 82.

L R

Figure 9-72 Problem 83.

•78 Consider a rocket that is in deep space and at rest relative to

an inertial reference frame. The rocket’s engine is to be fired for a

254 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

84 Figure 9-73 shows an overhead

view of two particles sliding at constant

velocity over a frictionless surface. The

particles have the same mass and the

same initial speed v4.00 m/s, and they

collide where their paths intersect. An

xaxis is arranged to bisect the angle be-

tween their incoming paths, such that

u40.0. The region to the right of the

collision is divided into four lettered

sections by the xaxis and four numbered dashed lines. In what re-

gion or along what line do the particles travel if the collision is (a)

completely inelastic, (b) elastic, and (c) inelastic? What are their fi-

nal speeds if the collision is (d) completely inelastic and (e) elastic?

85 Speed deamplifier. In Fig.

9-74, block 1 of mass m1slides along

an xaxis on a frictionless floor at

speed 4.00 m/s. Then it undergoes a

one-dimensional elastic collision

with stationary block 2 of mass m2

2.00m1. Next, block 2 undergoes a one-dimensional elastic collision

with stationary block 3 of mass m32.00m2. (a) What then is the

speed of block 3? Are (b) the speed, (c) the kinetic energy, and (d)

the momentum of block 3 greater than, less than, or the same as

the initial values for block 1?

origin with linear momentum (6.4 1023 kgm/s) . What are

the (a) magnitude and (b) direction of the linear momentum of the

daughter nucleus? (c) If the daughter nucleus has a mass of 5.8 

1026 kg,what is its kinetic energy?

91 A 75 kg man rides on a 39 kg cart moving at a velocity of 2.3 m/s.

He jumps off with zero horizontal velocity relative to the ground.

What is the resulting change in the cart’s velocity,including sign?

92 Two blocks of masses 1.0 kg and 3.0 kg are connected by a

spring and rest on a frictionless surface. They are given velocities

toward each other such that the 1.0 kg block travels initially at

1.7 m/s toward the center of mass, which remains at rest. What is

the initial speed of the other block?

93 A railroad freight car of mass 3.18 104kg collides

with a stationary caboose car.They couple together, and 27.0% of

the initial kinetic energy is transferred to thermal energy, sound,

vibrations, and so on. Find the mass of the caboose.

94 An old Chrysler with mass 2400 kg is moving along a straight

stretch of road at 80 km/h. It is followed by a Ford with mass 1600

kg moving at 60 km/h. How fast is the center of mass of the two

cars moving?

95 In the arrangement of Fig.9-21, billiard ball 1 moving at a

speed of 2.2 m/s undergoes a glancing collision with identical bil-

liard ball 2 that is at rest. After the collision, ball 2 moves at speed

1.1 m/s, at an angle of u260.What are (a) the magnitude and (b)

the direction of the velocity of ball 1 after the collision? (c) Do the

given data suggest the collision is elastic or inelastic?

96 A rocket is moving away from the solar system at a speed of

6.0 103m/s. It fires its engine, which ejects exhaust with a speed

of 3.0 103m/s relative to the rocket. The mass of the rocket at

this time is 4.0 104kg, and its acceleration is 2.0 m/s2. (a) What is

the thrust of the engine? (b) At what rate, in kilograms per second,

is exhaust ejected during the firing?

SSM

θθ

Figure 9-73 Problem 84.

321

Figure 9-74 Problem 85.

321

Figure 9-75 Problem 86.

Figure 9-76 Problem 97.

v0x

The velocity of ball 1 has magnitude v010 m/s and is directed at

the contact point of balls 1 and 2. After the collision, what are the

(a) speed and (b) direction of the velocity of ball 2, the (c) speed

and (d) direction of the velocity of ball 3, and the (e) speed and (f)

direction of the velocity of ball 1? (Hint: With friction absent, each

impulse is directed along the line connecting the centers of the col-

liding balls,normal to the colliding surfaces.)

98 A 0.15 kg ball hits a wall with a velocity of (5.00 m/s) (6.50

m/s) (4.00 m/s) . It rebounds from the wall with a velocity of

(2.00 m/s) (3.50 m/s) ( 3.20 m/s) . What are

(a) the change in the ball’s momentum, (b) the im-

pulse on the ball, and (c) the impulse on the wall?

99 In Fig. 9-77, two identical containers of sugar

are connected by a cord that passes over a friction-

less pulley. The cord and pulley have negligible

mass, each container and its sugar together have a

mass of 500 g, the centers of the containers are sepa-

rated by 50 mm, and the containers are held fixed at

the same height. What is the horizontal distance be-

tween the center of container 1 and the center

of mass of the two-container system (a) initially and

j

ˆi

ˆ

ˆi

97 The three balls in the

overhead view of Fig. 9-76 are

identical. Balls 2 and 3 touch

each other and are aligned per-

pendicular to the path of ball 1.

Figure 9-77

Problem 99.

1 2

sion with stationary block 2 of mass m20.500m1.Next, block 2 un-

dergoes a one-dimensional elastic collision with stationary block 3

of mass m30.500m2.(a) What then is the speed of block 3? Are (b)

the speed, (c) the kinetic energy, and (d) the momentum of block 3

greater than, less than, or the same as the initial values for block 1?

87 A ball having a mass of 150 g strikes a wall with a speed of

5.2 m/s and rebounds with only 50% of its initial kinetic energy. (a)

What is the speed of the ball immediately after rebounding? (b)

What is the magnitude of the impulse on the wall from the ball? (c) If

the ball is in contact with the wall for 7.6 ms, what is the magnitude of

the average force on the ball from the wall during this time interval?

88 A spacecraft is separated into two parts by detonating the ex-

plosive bolts that hold them together. The masses of the parts are

1200 kg and 1800 kg; the magnitude of the impulse on each part

from the bolts is 300 Ns. With what relative speed do the two

parts separate because of the detonation?

89 A 1400 kg car moving at 5.3 m/s is initially traveling

north along the positive direction of a yaxis. After completing a

90right-hand turn in 4.6 s, the inattentive operator drives into a

tree, which stops the car in 350 ms. In unit-vector notation, what is

the impulse on the car (a) due to the turn and (b) due to the colli-

sion? What is the magnitude of the average force that acts on the

car (c) during the turn and (d) during the collision? (e) What is the

direction of the average force during the turn?

90 A certain radioactive (parent) nucleus transforms to a dif-

ferent (daughter) nucleus by emitting an electron and a neutrino.

The parent nucleus was at rest at the origin of an xy coordinate sys-

tem. The electron moves away from the origin with linear momen-

tum (1.2 1022 kgm/s) ; the neutrino moves away from thei

ILW

SSM

86 Speed amplifier. In Fig. 9-75,

block 1 of mass m1slides along an x

axis on a frictionless floor with a

speed of v1i4.00 m/s. Then it under-

goes a one-dimensional elastic colli-



255

PROBLEMS

(b) after 20 g of sugar is transferred from container 1 to container

2? After the transfer and after the containers are released, (c) in

what direction and (d) at what acceleration magnitude does the

center of mass move?

100 In a game of pool, the cue ball strikes another ball of the

same mass and initially at rest. After the collision, the cue ball

moves at 3.50 m/s along a line making an angle of 22.0with the

cue ball’s original direction of motion, and the second ball has a

speed of 2.00 m/s. Find (a) the angle between the direction of mo-

tion of the second ball and the original direction of motion of the

cue ball and (b) the original speed of the cue ball. (c) Is kinetic en-

ergy (of the centers of mass, don’t consider the rotation) con-

served?

101 In Fig. 9-78, a 3.2 kg box of

running shoes slides on a horizontal

frictionless table and collides with a

2.0 kg box of ballet slippers initially

at rest on the edge of the table, at

height h0.40 m. The speed of the

3.2 kg box is 3.0 m/s just before the

collision. If the two boxes stick to-

gether because of packing tape on their

sides, what is their kinetic energy just before

they strike the floor?

102 In Fig. 9-79, an 80 kg man is on a lad-

der hanging from a balloon that has a total

mass of 320 kg (including the basket passen-

ger). The balloon is initially stationary rela-

tive to the ground. If the man on the ladder

begins to climb at 2.5 m/s relative to the lad-

der, (a) in what direction and (b) at what

speed does the balloon move? (c) If the man

then stops climbing, what is the speed of the

balloon?

103 In Fig. 9-80, block 1 of mass m16.6 kg

is at rest on a long frictionless table that is up

against a wall. Block 2 of mass m2is placed

between block 1 and the wall and sent sliding

to the left, toward block 1, with constant

speed v2i. Find the value of m2for which both

blocks move with the same velocity after block 2 has collided once

with block 1 and once with the wall.Assume all collisions are elastic

(the collision with the wall does not change the speed of block 2).

boat will initially touch the dock, as in Fig. 9-81; the boat can slide

through the water without significant resistance; both the car and

the boat can be approximated as uniform in their mass distribu-

tion. Determine what the width of the gap will be just as the car is

about to make the jump.

105 A 3.0 kg object moving at 8.0 m/s in the positive direc-

tion of an xaxis has a one-dimensional elastic collision with an ob-

ject of mass M, initially at rest. After the collision the object of

mass Mhas a velocity of 6.0 m/s in the positive direction of the

axis.What is mass M?

106 A 2140 kg railroad flatcar, which can move with negligible

friction, is motionless next to a platform. A 242 kg sumo wrestler

runs at 5.3 m/s along the platform (parallel to the track) and then

jumps onto the flatcar. What is the speed of the flatcar if he then

(a) stands on it, (b) runs at 5.3 m/s relative to it in his original direc-

tion, and (c) turns and runs at 5.3 m/s relative to the flatcar oppo-

site his original direction?

107 A 6100 kg rocket is set for vertical firing from the

ground. If the exhaust speed is 1200 m/s, how much gas must be

ejected each second if the thrust (a) is to equal the magnitude of

the gravitational force on the rocket and (b) is to give the rocket an

initial upward acceleration of 21 m/s2?

108 A 500.0 kg module is attached to a 400.0 kg shuttle craft,

which moves at 1000 m/s relative to the stationary main spaceship.

Then a small explosion sends the module backward with speed

100.0 m/s relative to the new speed of the shuttle craft. As meas-

ured by someone on the main spaceship, by what fraction did the

kinetic energy of the module and shuttle craft increase because of

the explosion?

109 (a) How far is the center of mass of the Earth–Moon

system from the center of Earth? (Appendix C gives the masses of

Earth and the Moon and the distance between the two.) (b) What

percentage of Earth’s radius is that distance?

110 A 140 g ball with speed 7.8 m/s strikes a wall perpendicu-

larly and rebounds in the opposite direction with the same speed.

The collision lasts 3.80 ms. What are the magnitudes of the (a) im-

pulse and (b) average force on the wall from the ball during the

elastic collision?

111 A rocket sled with a mass of 2900 kg moves at 250 m/s

on a set of rails. At a certain point, a scoop on the sled dips into a

trough of water located between the tracks and scoops water into

an empty tank on the sled. By applying the principle of conserva-

tion of linear momentum, determine the speed of the sled after

920 kg of water has been scooped up.Ignore any retarding force on

the scoop.

112 A pellet gun fires ten 2.0 g pellets per second with a

speed of 500 m/s.The pellets are stopped by a rigid wall.What are

(a) the magnitude of the momentum of each pellet, (b) the ki-

netic energy of each pellet, and (c) the magnitude of the average

force on the wall from the stream of pellets? (d) If each pellet is

in contact with the wall for 0.60 ms, what is the magnitude of the

average force on the wall from each pellet during contact? (e)

Why is this average force so different from the average force cal-

culated in (c)?

113 A railroad car moves under a grain elevator at a constant

speed of 3.20 m/s. Grain drops into the car at the rate of 540 kg/min.

What is the magnitude of the force needed to keep the car moving

at constant speed if friction is negligible?

SSM

Figure 9-78 Problem 101.

Figure 9-79

Problem 102.

v2i

Figure 9-80 Problem 103.

104 The script for an action movie calls for a small race car (of

mass 1500 kg and length 3.0 m) to accelerate along a flattop boat

(of mass 4000 kg and length 14 m), from one end of the boat to the

other, where the car will then jump

the gap between the boat and a

somewhat lower dock. You are the

technical advisor for the movie. The

A45678SF

Dock Boat

Figure 9-81 Problem 104.

256 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM

Figure 9-84 Problem 123.

114 Figure 9-82 shows a uniform square plate of edge length

6d6.0 m from which a square piece of edge length 2dhas been

removed.What are (a) the xcoordinate and (b) the ycoordinate of

the center of mass of the remaining piece?

rates the body into two parts, each of 4.0 kg, and increases the total

kinetic energy by 16 J. The forward part continues to move in the

original direction of motion. What are the speeds of (a) the rear

part and (b) the forward part?

121 An electron undergoes a one-dimensional elastic collision

with an initially stationary hydrogen atom.What percentage of the

electron’s initial kinetic energy is transferred to kinetic energy of

the hydrogen atom? (The mass of the hydrogen atom is 1840 times

the mass of the electron.)

122 A man (weighing 915 N) stands on a long railroad flatcar

(weighing 2415 N) as it rolls at 18.2 m/s in the positive direction of

an xaxis, with negligible friction.Then the man runs along the flat-

car in the negative xdirection at 4.00 m/s relative to the flatcar.

What is the resulting increase in the speed of the flatcar?

123 An unmanned space probe (of mass mand speed vrelative to

the Sun) approaches the planet Jupiter (of mass Mand speed VJrel-

ative to the Sun) as shown in Fig. 9-84. The spacecraft rounds the

planet and departs in the opposite direction. What is its speed (in

kilometers per second), relative to the Sun, after this slingshot en-

counter, which can be analyzed as a collision? Assume v10.5 km/s

and VJ13.0 km/s (the orbital speed of Jupiter).The mass of Jupiter

is very much greater than the mass of the spacecraft (Mm).

115 At time t0, force N acts on an

initially stationary particle of mass 2.00 103kg and force

N acts on an initially stationary particle of

mass 4.00 103kg. From time t0 to t2.00 ms, what are the

(a) magnitude and (b) angle (relative to the positive direction of

the xaxis) of the displacement of the center of mass of the two-

particle system? (c) What is the kinetic energy of the center of

mass at t2.00 ms?

116 Two particles Pand Qare released from rest 1.0 m apart.Phas

a mass of 0.10 kg, and Qa mass of 0.30 kg. Pand Qattract each other

with a constant force of 1.0 102N. No external forces act on the

system. (a) What is the speed of the center of mass of Pand Qwhen

the separation is 0.50 m? (b) At what distance from P’s original posi-

tion do the particles collide?

117 A collision occurs between a 2.00 kg particle traveling with

velocity and a 4.00 kg particle

traveling with velocity . The colli-

sion connects the two particles. What then is their velocity in (a)

unit-vector notation and as a (b) magnitude and (c) angle?

118 In the two-sphere arrangement of Fig. 9-20, assume that

sphere 1 has a mass of 50 g and an initial height of h19.0 cm, and

that sphere 2 has a mass of 85 g.After sphere 1 is released and col-

lides elastically with sphere 2, what height is reached by (a) sphere

1 and (b) sphere 2? After the next (elastic) collision, what height is

reached by (c) sphere 1 and (d) sphere 2? (Hint: Do not use

rounded-off values.)

119 In Fig. 9-83, block 1 slides along

an xaxis on a frictionless floor with a

speed of 0.75 m/s. When it reaches sta-

tionary block 2, the two blocks undergo

an elastic collision. The following table

gives the mass and length of the (uni-

form) blocks and also the locations of their centers at time t0.

Where is the center of mass of the two-block system located (a) at

t0, (b) when the two blocks first touch, and (c) at t4.0 s?

Block Mass (kg) Length (cm) Center at t0

1 0.25 5.0 x1.50 m

2 0.50 6.0 x0

120 A body is traveling at 2.0 m/s along the positive direction of

an xaxis; no net force acts on the body. An internal explosion sepa-

(2.00 m/s)j

2(6.00 m/s)i

ˆ

1(4.00 m/s)i

ˆ(5.00 m/s)j



2(2.00i

ˆ4.00j

ˆ)



1(4.00i

ˆ5.00j

ˆ)

SSM

3d3d

Figure 9-82 Problem 114.

124 A 0.550 kg ball falls directly down onto concrete, hitting it

with a speed of 12.0 m/s and rebounding directly upward with a

speed of 3.00 m/s. Extend a yaxis upward. In unit-vector notation,

what are (a) the change in the ball’s momentum, (b) the impulse

on the ball, and (c) the impulse on the concrete?

125 An atomic nucleus at rest at the origin of an xy coordinate

system transforms into three particles. Particle 1, mass 16.7 1027

kg,moves away from the origin at velocity (6.00 106m/s) ; particle

2, mass 8.35 1027 kg, moves away at velocity (8.00 106m/s) .

(a) In unit-vector notation, what is the linear momentum of the

third particle, mass 11.7 1027 kg? (b) How much kinetic energy

appears in this transformation?

126 Particle 1 of mass 200 g and speed 3.00 m/s undergoes a one-

dimensional collision with stationary particle 2 of mass 400 g.What

is the magnitude of the impulse on particle 1 if the collision is (a)

elastic and (b) completely inelastic?

127 During a lunar mission, it is necessary to increase the speed

of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative to

the Moon. The speed of the exhaust products from the rocket en-

gine is 1000 m/s relative to the spacecraft. What fraction of the

initial mass of the spacecraft must be burned and ejected to accom-

plish the speed increase?

128 A cue stick strikes a stationary pool ball, with an average

force of 32 N over a time of 14 ms. If the ball has mass 0.20 kg, what

speed does it have just after impact?

–1.50 m 0

1 2

Figure 9-83 Problem 119.

257

CHAPTER 10

Rotation

10-1ROTATIONAL VARIABLES

After reading this module, you should be able to . . .

10.01 Identify that if all parts of a body rotate around a fixed

axis locked together, the body is a rigid body. (This chapter

is about the motion of such bodies.)

10.02 Identify that the angular position of a rotating rigid body

is the angle that an internal reference line makes with a

fixed, external reference line.

10.03 Apply the relationship between angular displacement

and the initial and final angular positions.

10.04 Apply the relationship between average angular veloc-

ity, angular displacement, and the time interval for that dis-

placement.

10.05 Apply the relationship between average angular accel-

eration, change in angular velocity, and the time interval for

that change.

10.06 Identify that counterclockwise motion is in the positive

direction and clockwise motion is in the negative direction.

10.07 Given angular position as a function of time, calculate the

instantaneous angular velocity at any particular time and the

average angular velocity between any two particular times.

10.08 Given a graph of angular position versus time, deter-

mine the instantaneous angular velocity at a particular time

and the average angular velocity between any two particu-

lar times.

10.09 Identify instantaneous angular speed as the magnitude

of the instantaneous angular velocity.

10.10 Given angular velocity as a function of time, calculate

the instantaneous angular acceleration at any particular

time and the average angular acceleration between any

two particular times.

10.11 Given a graph of angular velocity versus time, deter-

mine the instantaneous angular acceleration at any partic-

ular time and the average angular acceleration between

any two particular times.

10.12 Calculate a body’s change in angular velocity by

integrating its angular acceleration function with respect

to time.

10.13 Calculate a body’s change in angular position by inte-

grating its angular velocity function with respect to time.

●To describe the rotation of a rigid body about a fixed axis,

called the rotation axis, we assume a reference line is fixed in

the body, perpendicular to that axis and rotating with the

body. We measure the angular position uof this line

relative to a fixed direction. When uis measured in radians,

(radian measure),

where sis the arc length of a circular path of radius rand

angle u.

●Radian measure is related to angle measure in revolutions

and degrees by

1 rev 3602prad.

●A body that rotates about a rotation axis, changing its angu-

lar position from u1to u2, undergoes an angular displacement

uu2u1,

where uis positive for counterclockwise rotation and nega-

tive for clockwise rotation.

●If a body rotates through an angular displacement uin a

time interval t, its average angular velocity vavg is

us

The (instantaneous) angular velocity vof the body is

Both vavg and vare vectors, with directions given by a

right-hand rule. They are positive for counterclockwise rota-

tion and negative for clockwise rotation. The magnitude of the

body’s angular velocity is the angular speed.

●If the angular velocity of a body changes from v1to v2in a

time interval tt2t1, the average angular acceleration

aavg of the body is

The (instantaneous) angular acceleration aof the body is

Both aavg and aare vectors.

adv

dt .

aavg v2v1

t2t1

v

t.

vdu

dt .

vavg u

t.

Key Ideas

Learning Objectives

258 CHAPTER 10 ROTATION

What Is Physics?

As we have discussed, one focus of physics is motion. However, so far we

have examined only the motion of translation, in which an object moves along

a straight or curved line, as in Fig. 10-1a. We now turn to the motion of rotation,

in which an object turns about an axis, as in Fig. 10-1b.

You see rotation in nearly every machine, you use it every time you open a

beverage can with a pull tab, and you pay to experience it every time you go to an

amusement park. Rotation is the key to many fun activities, such as hitting a long

drive in golf (the ball needs to rotate in order for the air to keep it aloft longer)

and throwing a curveball in baseball (the ball needs to rotate in order for the air

to push it left or right). Rotation is also the key to more serious matters, such as

metal failure in aging airplanes.

We begin our discussion of rotation by defining the variables for the

motion, just as we did for translation in Chapter 2. As we shall see, the vari-

ables for rotation are analogous to those for one-dimensional motion and, as

in Chapter 2, an important special situation is where the acceleration (here the

rotational acceleration) is constant. We shall also see that Newton’s second

law can be written for rotational motion, but we must use a new quantity

called torque instead of just force. Work and the work–kinetic energy

theorem can also be applied to rotational motion, but we must use a new quan-

tity called rotational inertia instead of just mass. In short, much of what we

have discussed so far can be applied to rotational motion with, perhaps, a few

changes.

Caution: In spite of this repetition of physics ideas, many students find this

and the next chapter very challenging. Instructors have a variety of reasons as

to why, but two reasons stand out: (1) There are a lot of symbols (with Greek

Figure 10-1 Figure skater Sasha Cohen in motion of (a) pure translation in a fixed

direction and (b) pure rotation about a vertical axis.

(b)

(a)

Mike Segar/Reuters/Landov LLC

Elsa/Getty Images, Inc.

259

10-1 ROTATIONAL VARIABLES

letters) to sort out. (2) Although you are very familiar with linear motion (you

can get across the room and down the road just fine), you are probably very

unfamiliar with rotation (and that is one reason why you are willing to pay so

much for amusement park rides). If a homework problem looks like a foreign

language to you, see if translating it into the one-dimensional linear motion of

Chapter 2 helps. For example, if you are to find, say, an angular distance, tem-

porarily delete the word angular and see if you can work the problem with the

Chapter 2 notation and ideas.

Rotational Variables

We wish to examine the rotation of a rigid body about a fixed axis.A rigid body is

a body that can rotate with all its parts locked together and without any change in

its shape. A fixed axis means that the rotation occurs about an axis that does not

move.Thus, we shall not examine an object like the Sun, because the parts of the

Sun (a ball of gas) are not locked together. We also shall not examine an object

like a bowling ball rolling along a lane, because the ball rotates about a moving

axis (the ball’s motion is a mixture of rotation and translation).

Figure 10-2 shows a rigid body of arbitrary shape in rotation about a fixed

axis, called the axis of rotation or the rotation axis. In pure rotation (angular

motion), every point of the body moves in a circle whose center lies on the axis of

rotation, and every point moves through the same angle during a particular time

interval. In pure translation (linear motion), every point of the body moves in a

straight line, and every point moves through the same linear distance during a

particular time interval.

We deal now—one at a time—with the angular equivalents of the linear

quantities position, displacement, velocity, and acceleration.

Angular Position

Figure 10-2 shows a reference line, fixed in the body, perpendicular to the rotation

axis and rotating with the body. The angular position of this line is the angle of

the line relative to a fixed direction, which we take as the zero angular position.

In Fig.10-3, the angular position uis measured relative to the positive direction of

the xaxis. From geometry, we know that uis given by

(radian measure). (10-1)

Here sis the length of a circular arc that extends from the xaxis (the zero angular

position) to the reference line, and ris the radius of the circle.

us

Figure 10-2 A rigid body of arbitrary shape in pure rotation about the zaxis of a coordinate

system. The position of the reference line with respect to the rigid body is arbitrary, but it is

perpendicular to the rotation axis. It is fixed in the body and rotates with the body.

Reference line

Rotation

axis

Body This reference line is part of the body

and perpendicular to the rotation axis.

We use it to measure the rotation of the

body relative to a fixed direction.

Figure 10-3 The rotating rigid body of

Fig. 10-2 in cross section, viewed from

above. The plane of the cross section is

perpendicular to the rotation axis, which

now extends out of the page, toward you.

In this position of the body, the reference

line makes an angle uwith the xaxis.

Reference

line

Rotation

axis

The body has rotated

counterclockwise

by angle . This is the

positive direction.

This dot means that

the rotation axis is

out toward you.

An angle defined in this way is measured in radians (rad) rather than in

revolutions (rev) or degrees. The radian, being the ratio of two lengths, is a

pure number and thus has no dimension. Because the circumference of a circle of

radius ris 2pr, there are 2pradians in a complete circle:

(10-2)

and thus 1 rad 57.30.159 rev. (10-3)

We do not reset uto zero with each complete rotation of the reference line about

the rotation axis. If the reference line completes two revolutions from the zero

angular position, then the angular position uof the line is u4prad.

For pure translation along an xaxis, we can know all there is to know

about a moving body if we know x(t), its position as a function of time.

Similarly, for pure rotation, we can know all there is to know about a rotating

body if we know u(t), the angular position of the body’s reference line as a

function of time.

Angular Displacement

If the body of Fig. 10-3 rotates about the rotation axis as in Fig. 10-4, changing the

angular position of the reference line from u1to u2, the body undergoes an

angular displacement ugiven by

uu2u1. (10-4)

This definition of angular displacement holds not only for the rigid body as a

whole but also for every particle within that body.

Clocks Are Negative. If a body is in translational motion along an xaxis, its

displacement xis either positive or negative, depending on whether the body is

moving in the positive or negative direction of the axis. Similarly, the angular dis-

placement uof a rotating body is either positive or negative, according to the

following rule:

1 rev 360 2pr

r2p rad,

260 CHAPTER 10 ROTATION

An angular displacement in the counterclockwise direction is positive, and one in

the clockwise direction is negative.

Checkpoint 1

A disk can rotate about its central axis like a merry-go-round.Which of the following

pairs of values for its initial and final angular positions,respectively, give a negative

angular displacement: (a) 3 rad, 5 rad, (b) 3 rad, 7 rad, (c) 7 rad, 3 rad?

The phrase “clocks are negative” can help you remember this rule (they certainly

are negative when their alarms sound off early in the morning).

Angular Velocity

Suppose that our rotating body is at angular position u1at time t1and at

angular position u2at time t2as in Fig. 10-4. We define the average angular velocity

of the body in the time interval tfrom t1to t2to be

(10-5)

where uis the angular displacement during t(vis the lowercase omega).

vavg u2u1

t2t1

u

t,

261

10-1 ROTATIONAL VARIABLES

Figure 10-4 The reference line of the rigid body of Figs. 10-2 and 10-3 is at angular position

u1at time t1and at angular position u2at a later time t2. The quantity u(u2u1) is the

angular displacement that occurs during the interval t(t2t1).The body itself is not

shown.

Rotation axis

At t2

At t1

Reference line

This change in the angle of the reference line

(which is part of the body) is equal to the angular

displacement of the body itself during this

time interval.

The (instantaneous) angular velocity v, with which we shall be most con-

cerned, is the limit of the ratio in Eq. 10-5 as tapproaches zero.Thus,

(10-6)

If we know u(t), we can find the angular velocity vby differentiation.

Equations 10-5 and 10-6 hold not only for the rotating rigid body as a whole

but also for every particle of that body because the particles are all locked

together. The unit of angular velocity is commonly the radian per second (rad/s)

or the revolution per second (rev/s). Another measure of angular velocity was

used during at least the first three decades of rock: Music was produced by vinyl

(phonograph) records that were played on turntables at “ ” or “45 rpm,”

meaning at or 45 rev/min.

If a particle moves in translation along an xaxis, its linear velocity vis either

positive or negative, depending on its direction along the axis. Similarly, the angu-

lar velocity vof a rotating rigid body is either positive or negative, depending on

whether the body is rotating counterclockwise (positive) or clockwise (negative).

(“Clocks are negative” still works.) The magnitude of an angular velocity is called

the angular speed, which is also represented with v.

Angular Acceleration

If the angular velocity of a rotating body is not constant, then the body has an an-

gular acceleration. Let v2and v1be its angular velocities at times t2and t1,

respectively.The average angular acceleration of the rotating body in the interval

from t1to t2is defined as

(10-7)

in which vis the change in the angular velocity that occurs during the time

interval t. The (instantaneous) angular acceleration a, with which we shall be

most concerned, is the limit of this quantity as tapproaches zero.Thus,

(10-8)

As the name suggests, this is the angular acceleration of the body at a given in-

stant. Equations 10-7 and 10-8 also hold for every particle of that body. The unit of

angular acceleration is commonly the radian per second-squared (rad/s2) or the

revolution per second-squared (rev/s2).

alim

t:0

v

tdv

dt .

aavg v2v1

t2t1

v

t,

331

3 rev/min

331

3 rpm

vlim

t:0

u

tdu

dt .

262 CHAPTER 10 ROTATION

Calculations: To sketch the disk and its reference line at a

particular time, we need to determine ufor that time. To do

so, we substitute the time into Eq. 10-9. For t2.0 s, we get

This means that at t2.0 s the reference line on the disk

is rotated counterclockwise from the zero position by angle

1.2 rad 69(counterclockwise because uis positive). Sketch

1 in Fig.10-5bshows this position of the reference line.

Similarly, for t0, we find u1.00 rad 57,

which means that the reference line is rotated clockwise

from the zero angular position by 1.0 rad, or 57, as shown

in sketch 3. For t4.0 s, we find u0.60 rad 34

(sketch 5). Drawing sketches for when the curve crosses

the taxis is easy, because then u0 and the reference line

is momentarily aligned with the zero angular position

(sketches 2 and 4).

(b) At what time tmin does u(t) reach the minimum

value shown in Fig.10-5b? What is that minimum value?

1.2 rad 1.2 rad 360



rad 69.

u1.00 (0.600)(2.0) (0.250)(2.0)2

Sample Problem 10.01 Angular velocity derived from angular position

The disk in Fig. 10-5ais rotating about its central axis like a

merry-go-round. The angular position u(t) of a reference

line on the disk is given by

u1.00 0.600t0.250t2, (10-9)

with tin seconds, uin radians, and the zero angular position

as indicated in the figure. (If you like, you can translate all

this into Chapter 2 notation by momentarily dropping the

word “angular” from “angular position” and replacing the

symbol uwith the symbol x.What you then have is an equa-

tion that gives the position as a function of time, for the one-

dimensional motion of Chapter 2.)

(a) Graph the angular position of the disk versus time

from t3.0 s to t5.4 s. Sketch the disk and its angular

position reference line at t2.0 s, 0 s, and 4.0 s, and

when the curve crosses the taxis.

KEY IDEA

The angular position of the disk is the angular position

u(t) of its reference line, which is given by Eq. 10-9 as a function

of time t. So we graph Eq. 10-9; the result is shown in Fig. 10-5b.

Zero

angular

position

Reference

line

Rotation axis

(a)

(b)

–2 024 6

(rad)

(1) (2) (3) (4) (5)

t(s)

–2

The angular position

of the disk is the angle

between these two lines.

Now, the disk is

at a zero angle.

At t=−2 s, the disk

is at a positive

(counterclockwise)

angle. So, a positive

value is plotted.

This is a plot of the angle

of the disk versus time.

Now, it is at a

negative (clockwise)

angle. So, a negative

value is plotted.

It has reversed

its rotation and

is again at a

zero angle.

Now, it is

back at a

positive

angle.

Figure 10-5 (a) A rotating disk. (b) A plot of the disk’s angular position u(t). Five sketches indicate the angular position of the refer-

ence line on the disk for five points on the curve. (c) A plot of the disk’s angular velocity v(t). Positive values of vcorrespond to

counterclockwise rotation, and negative values to clockwise rotation.

263

10-1 ROTATIONAL VARIABLES

t3.0 s to t6.0 s. Sketch the disk and indicate the direc-

tion of turning and the sign of vat t2.0 s,4.0 s,and tmin.

KEY IDEA

From Eq. 10-6, the angular velocity vis equal to du/dt as

given in Eq. 10-10. So, we have

v0.600 0.500t. (10-11)

The graph of this function v(t) is shown in Fig. 10-5c.

Because the function is linear, the plot is a straight line. The

slope is 0.500 rad/s2and the intercept with the vertical axis

(not shown) is 0.600 rad/s.

Calculations: To sketch the disk at t2.0 s, we substitute

that value into Eq. 10-11, obtaining

v1.6 rad/s. (Answer)

The minus sign here tells us that at t2.0 s, the disk is

turning clockwise (as indicated by the left-hand sketch in

Fig. 10-5c).

Substituting t4.0 s into Eq. 10-11 gives us

v1.4 rad/s. (Answer)

The implied plus sign tells us that now the disk is turning

counterclockwise (the right-hand sketch in Fig.10-5c).

For tmin, we already know that du/dt 0. So, we must

also have v0. That is, the disk momentarily stops when

the reference line reaches the minimum value of uin

Fig. 10-5b, as suggested by the center sketch in Fig. 10-5c. On

the graph of v versus tin Fig. 10-5c, this momentary stop is

the zero point where the plot changes from the negative

clockwise motion to the positive counterclockwise motion.

(d) Use the results in parts (a) through (c) to describe the

motion of the disk from t3.0 s to t6.0 s.

Description: When we first observe the disk at t3.0 s, it

has a positive angular position and is turning clockwise but

slowing. It stops at angular position u1.36 rad and then

begins to turn counterclockwise, with its angular position

eventually becoming positive again.

KEY IDEA

To find the extreme value (here the minimum) of a function,

we take the first derivative of the function and set the result

to zero.

Calculations: The first derivative of u(t) is

(10-10)

Setting this to zero and solving for tgive us the time at

which u(t) is minimum:

tmin 1.20 s. (Answer)

To get the minimum value of u, we next substitute tmin into

Eq. 10-9, finding

u1.36 rad 77.9. (Answer)

This minimum of u(t) (the bottom of the curve in Fig. 10-5b)

corresponds to the maximum clockwise rotation of the disk

from the zero angular position, somewhat more than is

shown in sketch 3.

dt 0.600 0.500t.

(c)

–2 –2 0 2 4 6

(rad/s)

t(s)

negative

zero

positive

This is a plot of the angular

velocity of the disk versus time.

The angular velocity is

initially negative and slowing,

then momentarily zero during

reversal, and then positive and

increasing.

Additional examples, video, and

practice available at WileyPLUS

264 CHAPTER 10 ROTATION

Are Angular Quantities Vectors?

We can describe the position, velocity, and acceleration of a single particle by

means of vectors. If the particle is confined to a straight line, however, we do not

really need vector notation. Such a particle has only two directions available to it,

and we can indicate these directions with plus and minus signs.

In the same way, a rigid body rotating about a fixed axis can rotate only

clockwise or counterclockwise as seen along the axis, and again we can select

between the two directions by means of plus and minus signs.The question arises:

“Can we treat the angular displacement, velocity, and acceleration of a rotating

body as vectors?” The answer is a qualified “yes” (see the caution below, in con-

nection with angular displacements).

Angular Velocities. Consider the angular velocity. Figure 10-6ashows a

vinyl record rotating on a turntable. The record has a constant angular speed

in the clockwise direction. We can represent its angular ve-

locity as a vector pointing along the axis of rotation, as in Fig. 10-6b. Here’s

how: We choose the length of this vector according to some convenient scale,

for example, with 1 cm corresponding to 10 rev/min.Then we establish a direc-

tion for the vector by using a right-hand rule, as Fig. 10-6cshows: Curl your

right hand about the rotating record, your fingers pointing in the direction of

rotation. Your extended thumb will then point in the direction of the angular

velocity vector. If the record were to rotate in the opposite sense, the right-

(331

3 rev/min)

To evaluate the constant of integration C, we note that v

5 rad/s at t0. Substituting these values in our expression

for vyields

so C5 rad/s.Then

. (Answer)

(b) Obtain an expression for the angular position u(t) of the

top.

KEY IDEA

By definition, v(t) is the derivative of u(t) with respect to

time. Therefore, we can find u(t) by integrating v(t) with

respect to time.

Calculations: Since Eq. 10-6 tells us that

duv dt,

we can write

(Answer)

where Chas been evaluated by noting that u2 rad at t0.

1

4t52

3t35t2,

1

4t52

3t35tC

uv

dt  (5

4t42t25) dt

v5

4t42t25

5 rad/s 0 0 C

Sample Problem 10.02 Angular velocity derived from angular acceleration

A child’s top is spun with angular acceleration

with tin seconds and ain radians per second-squared. At

t0, the top has angular velocity 5 rad/s, and a reference

line on it is at angular position u2 rad.

(a) Obtain an expression for the angular velocity v(t) of the

top.That is, find an expression that explicitly indicates how the

angular velocity depends on time. (We can tell that there is

such a dependence because the top is undergoing an angular

acceleration,which means that its angular velocity is changing.)

KEY IDEA

By definition, a(t) is the derivative of v(t) with respect to time.

Thus, we can find v(t) by integrating a(t) with respect to time.

Calculations: Equation 10-8 tells us

so .

From this we find

.v(5t34t)dt 5

4t44

2t2C

dva

dvadt

a5t34t

Additional examples, video, and practice available at WileyPLUS

265

10-1 ROTATIONAL VARIABLES

hand rule would tell you that the angular velocity vector then points in the op-

posite direction.

It is not easy to get used to representing angular quantities as vectors. We in-

stinctively expect that something should be moving along the direction of a vec-

tor. That is not the case here. Instead, something (the rigid body) is rotating

around the direction of the vector. In the world of pure rotation, a vector defines

an axis of rotation, not a direction in which something moves. Nonetheless, the

vector also defines the motion. Furthermore, it obeys all the rules for vector

manipulation discussed in Chapter 3. The angular acceleration is another

vector,and it too obeys those rules.

In this chapter we consider only rotations that are about a fixed axis. For such

situations, we need not consider vectors—we can represent angular velocity with

vand angular acceleration with a, and we can indicate direction with an implied

plus sign for counterclockwise or an explicit minus sign for clockwise.

Angular Displacements.Now for the caution: Angular displacements

(unless they are very small) cannot be treated as vectors. Why not? We can cer-

tainly give them both magnitude and direction, as we did for the angular veloc-

ity vector in Fig. 10-6. However, to be represented as a vector, a quantity must

also obey the rules of vector addition, one of which says that if you add two

vectors, the order in which you add them does not matter. Angular displace-

ments fail this test.

Figure 10-7 gives an example. An initially horizontal book is given two

90angular displacements, first in the order of Fig. 10-7aand then in the order

of Fig. 10-7b.Although the two angular displacements are identical, their order

is not, and the book ends up with different orientations. Here’s another exam-

ple. Hold your right arm downward, palm toward your thigh. Keeping your

wrist rigid, (1) lift the arm forward until it is horizontal, (2) move it horizon-

tally until it points toward the right, and (3) then bring it down to your side.

Your palm faces forward. If you start over, but reverse the steps, which way

does your palm end up facing? From either example, we must conclude that

the addition of two angular displacements depends on their order and they

cannot be vectors.

Figure 10-6 (a) A record rotating about a vertical axis that coincides with the axis of the

spindle. (b) The angular velocity of the rotating record can be represented by the vector

, lying along the axis and pointing down, as shown. (c) We establish the direction of the

angular velocity vector as downward by using a right-hand rule. When the fingers of the

right hand curl around the record and point the way it is moving, the extended thumb

points in the direction of .v

zz z

(a) (b) (c)

Axis Axis Axis

Spindle

This right-hand rule

establishes the

direction of the

angular velocity

vector.

Figure 10-7 (a) From its initial position, at

the top, the book is given two successive

90rotations, first about the (horizontal)

xaxis and then about the (vertical) yaxis.

(b) The book is given the same rotations,

but in the reverse order.

PHYSICS

(a) (b)

PHYSICS

zThe order of the

rotations makes

a big difference

in the result.

266 CHAPTER 10 ROTATION

Rotation with Constant Angular Acceleration

In pure translation, motion with a constant linear acceleration (for example, that

of a falling body) is an important special case. In Table 2-1, we displayed a series

of equations that hold for such motion.

In pure rotation, the case of constant angular acceleration is also important,

and a parallel set of equations holds for this case also. We shall not derive them

here, but simply write them from the corresponding linear equations, substituting

equivalent angular quantities for the linear ones.This is done in Table 10-1, which

lists both sets of equations (Eqs. 2-11 and 2-15 to 2-18; 10-12 to 10-16).

Recall that Eqs. 2-11 and 2-15 are basic equations for constant linear

acceleration—the other equations in the Linear list can be derived from them.

Similarly, Eqs. 10-12 and 10-13 are the basic equations for constant angular

acceleration, and the other equations in the Angular list can be derived from

them.To solve a simple problem involving constant angular acceleration, you can

usually use an equation from the Angular list (if you have the list). Choose

an equation for which the only unknown variable will be the variable requested

in the problem.A better plan is to remember only Eqs. 10-12 and 10-13, and then

solve them as simultaneous equations whenever needed.

10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION

After reading this module, you should be able to . . .

10.14 For constant angular acceleration, apply the relation-

ships between angular position, angular displacement,

Key Idea

●Constant angular acceleration (aconstant) is an important special case of rotational motion. The appropriate kinematic

equations are

vv0at,

uu0vt1

2at2.

uu01

2(v0v)t,

v2v0

22a(uu0),

uu0v0t1

2at2,

Learning Objective

angular velocity, angular acceleration, and elapsed time

(Table 10-1).

Table 10-1 Equations of Motion for Constant Linear Acceleration and for Constant Angular Acceleration

Equation Linear Missing Angular Equation

Number Equation Variable Equation Number

(2-11) vv0at x x0uu0vv0at(10-12)

(2-15) vv(10-13)

(2-16) tt (10-14)

(2-17) aa(10-15)

(2-18) v0v0(10-16)uu0vt1

2at2

xx0vt 1

2at2

uu01

2(v0v)tx x01

2(v0v)t

v2v0

22a(uu0)v2v0

22a(xx0)

uu0v0t1

2at2

xx0v0t1

2at2

Checkpoint 2

In four situations, a rotating body has angular position u(t) given by (a) u3t4,

(b) u5t34t26, (c) u2/t24/t,and (d) u5t23.To which situations do

the angular equations of Table 10-1 apply?

267

10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION

(We converted 5.0 rev to 10prad to keep the units consis-

tent.) Solving this quadratic equation for t,we find

t32 s. (Answer)

Now notice something a bit strange. We first see the wheel

when it is rotating in the negative direction and through the

u0 orientation.Yet, we just found out that 32 s later it is at

the positive orientation of u5.0 rev. What happened in

that time interval so that it could be at a positive orientation?

(b) Describe the grindstone’s rotation between t0 and

t32 s.

Description: The wheel is initially rotating in the negative

(clockwise) direction with angular velocity v04.6 rad/s,

but its angular acceleration ais positive.This initial opposi-

tion of the signs of angular velocity and angular accelera-

tion means that the wheel slows in its rotation in the nega-

tive direction, stops, and then reverses to rotate in the

positive direction. After the reference line comes back

through its initial orientation of u0, the wheel turns an

additional 5.0 rev by time t32 s.

Calculation: We again go to the table of equations for con-

stant angular acceleration, and again we need an equation

that contains only the desired unknown variable t. However,

now the equation must also contain the variable v, so that we

can set it to 0 and then solve for the corresponding time t. We

choose Eq. 10-12,which yields

(Answer)tvv0

a0(4.6 rad/s)

0.35 rad/s213 s.

Sample Problem 10.03 Constant angular acceleration, grindstone

A grindstone (Fig. 10-8) rotates at constant angular acceler-

ation a0.35 rad/s2. At time t0, it has an angular velocity

of v04.6 rad/s and a reference line on it is horizontal, at

the angular position u00.

(a) At what time after t0 is the reference line at the

angular position u5.0 rev?

KEY IDEA

The angular acceleration is constant, so we can use the rota-

tion equations of Table 10-1.We choose Eq.10-13,

because the only unknown variable it contains is the desired

time t.

Calculations: Substituting known values and setting u00

and u5.0 rev 10prad give us

.10p rad (4.6 rad/s)t1

2 (0.35 rad/s2)t2

uu0v0t1

2at2

Figure 10-8 A grindstone.At t0 the reference line (which we

imagine to be marked on the stone) is horizontal.

Axis

Reference

line

Zero angular

position

We measure rotation by using

this reference line.

Clockwise = negative

Counterclockwise = positive

rad/s, the angular displacement is uu020.0 rev, and the

angular velocity at the end of that displacement is v2.00

rad/s. In addition to the angular acceleration athat we want,

both basic equations also contain time t, which we do not

necessarily want.

To eliminate the unknown t, we use Eq. 10-12 to write

which we then substitute into Eq. 10-13 to write

Solving for a, substituting known data, and converting

20 rev to 125.7 rad, we find

(Answer)0.0301 rad/s2.

av2v0

2(uu0)(2.00 rad/s)2(3.40 rad/s)2

2(125.7 rad)

uu0v0



vv0



1



vv0



tvv0

Sample Problem 10.04 Constant angular acceleration, riding a Rotor

While you are operating a Rotor (a large, vertical, rotating

cylinder found in amusement parks), you spot a passenger in

acute distress and decrease the angular velocity of the cylin-

der from 3.40 rad/s to 2.00 rad/s in 20.0 rev, at constant angu-

lar acceleration. (The passenger is obviously more of a “trans-

lation person” than a “rotation person.”)

(a) What is the constant angular acceleration during this

decrease in angular speed?

KEY IDEA

Because the cylinder’s angular acceleration is constant, we

can relate it to the angular velocity and angular displacement

via the basic equations for constant angular acceleration

(Eqs. 10-12 and 10-13).

Calculations: Let’s first do a quick check to see if we can solve

the basic equations. The initial angular velocity is v03.40

268 CHAPTER 10 ROTATION

Relating the Linear and Angular Variables

In Module 4-5,we discussed uniform circular motion,in which a particle travels at con-

stant linear speed valong a circle and around an axis of rotation. When a rigid body,

such as a merry-go-round,rotates around an axis,each particle in the body moves in its

own circle around that axis. Since the body is rigid, all the particles make one revolu-

tion in the same amount of time;that is,they all have the same angular speed v.

However, the farther a particle is from the axis, the greater the circumference

of its circle is, and so the faster its linear speed vmust be.You can notice this on a

merry-go-round. You turn with the same angular speed vregardless of your dis-

tance from the center, but your linear speed vincreases noticeably if you move to

the outside edge of the merry-go-round.

We often need to relate the linear variables s,v, and afor a particular point in

a rotating body to the angular variables u,v, and afor that body. The two sets of

variables are related by r, the perpendicular distance of the point from the

rotation axis. This perpendicular distance is the distance between the point and

the rotation axis, measured along a perpendicular to the axis. It is also the radius r

of the circle traveled by the point around the axis of rotation.

(b) How much time did the speed decrease take?

Calculation: Now that we know a, we can use Eq. 10-12 to

solve for t:(Answer)46.5 s.

tvv0

a2.00 rad/s 3.40 rad/s

0.0301 rad/s2

10-3 RELATING THE LINEAR AND ANGULAR VARIABLES

After reading this module, you should be able to . . .

10.15 For a rigid body rotating about a fixed axis, relate the angular

variables of the body (angular position, angular velocity, and an-

gular acceleration) and the linear variables of a particle on the

body (position, velocity, and acceleration) at any given radius.

10.16 Distinguish between tangential acceleration and radial

acceleration, and draw a vector for each in a sketch of a

particle on a body rotating about an axis, for both an in-

crease in angular speed and a decrease.

●A point in a rigid rotating body, at a perpendicular distance

rfrom the rotation axis, moves in a circle with radius r. If the

body rotates through an angle u, the point moves along an

arc with length sgiven by

sur(radian measure),

where uis in radians.

●The linear velocity of the point is tangent to the circle; the

point’s linear speed vis given by

vvr(radian measure),

where vis the angular speed (in radians per second) of the body,

and thus also the point.

●The linear acceleration of the point has both tangential

and radial components. The tangential component is

atar(radian measure),

where ais the magnitude of the angular acceleration (in radi-

ans per second-squared) of the body. The radial component

of is

(radian measure).

●If the point moves in uniform circular motion, the period Tof

the motion for the point and the body is

(radian measure).T2pr

v2p

arv2

rv2r

Learning Objectives

Key Ideas

Additional examples, video, and practice available at WileyPLUS

269

10-3 RELATING THE LINEAR AND ANGULAR VARIABLES

The Position

If a reference line on a rigid body rotates through an angle u, a point within the

body at a position rfrom the rotation axis moves a distance salong a circular arc,

where sis given by Eq. 10-1:

sur(radian measure). (10-17)

This is the first of our linear–angular relations. Caution: The angle uhere must be

measured in radians because Eq. 10-17 is itself the definition of angular measure

in radians.

The Speed

Differentiating Eq. 10-17 with respect to time—with rheld constant—leads to

However, ds/dt is the linear speed (the magnitude of the linear velocity) of the

point in question, and du/dt is the angular speed vof the rotating body. So

vvr(radian measure). (10-18)

Caution: The angular speed vmust be expressed in radian measure.

Equation 10-18 tells us that since all points within the rigid body have the

same angular speed v, points with greater radius rhave greater linear speed v.

Figure 10-9areminds us that the linear velocity is always tangent to the circular

path of the point in question.

If the angular speed vof the rigid body is constant, then Eq. 10-18 tells

us that the linear speed vof any point within it is also constant. Thus, each point

within the body undergoes uniform circular motion. The period of revolution T

for the motion of each point and for the rigid body itself is given by Eq. 4-35:

. (10-19)

This equation tells us that the time for one revolution is the distance 2prtraveled

in one revolution divided by the speed at which that distance is traveled.

Substituting for vfrom Eq. 10-18 and canceling r, we find also that

(radian measure). (10-20)

This equivalent equation says that the time for one revolution is the angular dis-

tance 2prad traveled in one revolution divided by the angular speed (or rate) at

which that angle is traveled.

The Acceleration

Differentiating Eq. 10-18 with respect to time—again with rheld constant—

leads to

(10-21)

Here we run up against a complication. In Eq. 10-21, dv/dt represents only the

part of the linear acceleration that is responsible for changes in the magnitude v

of the linear velocity . Like , that part of the linear acceleration is tangent to

the path of the point in question.We call it the tangential component atof the lin-

ear acceleration of the point, and we write

atar(radian measure), (10-22)

dt dv

dt r.

T2p

T2pr

dt du

dt r.

Figure 10-9 The rotating rigid body of Fig. 10-2,

shown in cross section viewed from above.

Every point of the body (such as P) moves

in a circle around the rotation axis. (a)The

linear velocity of every point is tangent to

the circle in which the point moves. (b) The

linear acceleration of the point has (in

general) two components: tangential atand

radial ar.

Rotation

axis

Circle

traveled by P

(a)

The velocity vector is

always tangent to this

circle around the

rotation axis.

(b)

Rotation

axis

The acceleration always

has a radial (centripetal)

component and may have

a tangential component.

270 CHAPTER 10 ROTATION

where adv/dt. Caution: The angular acceleration ain Eq. 10-22 must be

expressed in radian measure.

In addition, as Eq. 4-34 tells us, a particle (or point) moving in a circular path

has a radial component of linear acceleration, arv2/r(directed radially inward),

that is responsible for changes in the direction of the linear velocity . By substi-

tuting for vfrom Eq. 10-18, we can write this component as

(radian measure). (10-23)

Thus, as Fig. 10-9bshows, the linear acceleration of a point on a rotating rigid

body has, in general, two components. The radially inward component ar(given

by Eq. 10-23) is present whenever the angular velocity of the body is not zero.

The tangential component at(given by Eq. 10-22) is present whenever the angu-

lar acceleration is not zero.

arv2

rv2r

Checkpoint 3

A cockroach rides the rim of a rotating merry-go-round. If the angular speed of this

system (merry-go-round cockroach) is constant,does the cockroach have (a) radial

acceleration and (b) tangential acceleration? If vis decreasing,does the cockroach

have (c) radial acceleration and (d) tangential acceleration?

and radial accelerations are the (perpendicular) compo-

nents of the (full) acceleration .

Calculations: Let’s go through the steps. We first find the

angular velocity by taking the time derivative of the given

angular position function and then substituting the given

time of t2.20 s:

v(ct3)3ct2(10-25)

3(6.39 102rad/s3)(2.20 s)2

0.928 rad/s. (Answer)

From Eq. 10-18, the linear speed just then is

vvr3ct2r(10-26)

3(6.39 102rad/s3)(2.20 s)2(33.1 m)

30.7 m/s. (Answer)

dt d

Sample Problem 10.05 Designing The Giant Ring, a large-scale amusement park ride

We are given the job of designing a large horizontal ring

that will rotate around a vertical axis and that will have a ra-

dius of r33.1 m (matching that of Beijing’s The Great

Observation Wheel, the largest Ferris wheel in the world).

Passengers will enter through a door in the outer wall of the

ring and then stand next to that wall (Fig.10-10a).We decide

that for the time interval t0 to t2.30 s, the angular posi-

tion u(t) of a reference line on the ring will be given by

uct3, (10-24)

with c6.39 102rad/s3. After t2.30 s, the angular

speed will be held constant until the end of the ride. Once

the ring begins to rotate, the floor of the ring will drop away

from the riders but the riders will not fall—indeed, they feel

as though they are pinned to the wall. For the time t2.20 s,

let’s determine a rider’s angular speed v, linear speed v,an-

gular acceleration a, tangential acceleration at, radial accel-

eration ar, and acceleration .

KEY IDEAS

(1) The angular speed vis given by Eq. 10-6 (vdu/dt).

(2) The linear speed v(along the circular path) is related to

the angular speed (around the rotation axis) by Eq. 10-18

(vvr). (3) The angular acceleration ais given by Eq. 10-8

(adv/dt). (4) The tangential acceleration at(along the cir-

cular path) is related to the angular acceleration (around

the rotation axis) by Eq. 10-22 (atar). (5) The radial accel-

eration aris given Eq. 10-23 (arv2r). (6) The tangential

(b)(a)

Figure 10-10 (a) Overhead view of

a passenger ready to ride The

Giant Ring. (b) The radial and

tangential acceleration compo-

nents of the (full) acceleration.

271

10-4 KINETIC ENERGY OF ROTATION

Additional examples, video, and practice available at WileyPLUS

The radial and tangential accelerations are perpendicu-

lar to each other and form the components of the rider’s

acceleration (Fig. 10-10b). The magnitude of is given by

a(10-29)

39.9 m/s2, (Answer)

or 4.1g(which is really exciting!). All these values are

acceptable.

To find the orientation of , we can calculate the angle u

shown in Fig.10-10b:

tan u

However, instead of substituting our numerical results, let’s

use the algebraic results from Eqs. 10-27 and 10-28:

utan1. (10-30)

The big advantage of solving for the angle algebraically is that

we can then see that the angle (1) does not depend on the

ring’s radius and (2) decreases as tgoes from 0 to 2.20 s. That

is, the acceleration vector swings toward being radially in-

ward because the radial acceleration (which depends on t4)

quickly dominates over the tangential acceleration (which

depends on only t).At our given time t2.20 s,we have

u . (Answer)tan12

3(6.39 102 rad/s3)(2.20 s)344.4



6ctr

9c2t4r



tan1



3ct3





2(28.49 m/s2)2(27.91 m/s2)2

2a2

ra2

Although this is fast (111 km/h or 68.7 mi/h), such speeds are

common in amusement parks and not alarming because (as

mentioned in Chapter 2) your body reacts to accelerations but

not to velocities. (It is an accelerometer, not a speedometer.)

From Eq. 10-26 we see that the linear speed is increasing as the

square of the time (but this increase will cut off at t2.30 s).

Next, let’s tackle the angular acceleration by taking the

time derivative of Eq. 10-25:

a(3ct2)6ct

6(6.39 102rad/s3)(2.20 s) 0.843 rad/s2. (Answer)

The tangential acceleration then follows from Eq. 10-22:

atar6ctr (10-27)

6(6.39 102rad/s3)(2.20 s)(33.1 m)

27.91 m/s227.9 m/s2, (Answer)

or 2.8g(which is reasonable and a bit exciting). Equation

10-27 tells us that the tangential acceleration is increasing

with time (but it will cut off at t2.30 s). From Eq. 10-23,

we write the radial acceleration as

arv2r.

Substituting from Eq. 10-25 leads us to

ar(3ct2)2r9c2t4r(10-28)

9(6.39 102rad/s3)2(2.20 s)4(33.1 m)

28.49 m/s228.5 m/s2, (Answer)

or 2.9g(which is also reasonable and a bit exciting).



dt d

10-4 KINETIC ENERGY OF ROTATION

After reading this module, you should be able to . . .

10.17 Find the rotational inertia of a particle about a point.

10.18 Find the total rotational inertia of many particles moving

around the same fixed axis.

10.19 Calculate the rotational kinetic energy of a

body in terms of its rotational inertia and its angular

speed.

●The kinetic energy Kof a rigid body rotating about a fixed

axis is given by

(radian measure),K1

2Iv2

in which Iis the rotational inertia of the body, defined as

for a system of discrete particles.

I



miri

Learning Objectives

Key Idea

Kinetic Energy of Rotation

The rapidly rotating blade of a table saw certainly has kinetic energy due to that

rotation. How can we express the energy? We cannot apply the familiar formula

to the saw as a whole because that would give us the kinetic energy

only of the saw’s center of mass, which is zero.

K1

2mv2

272 CHAPTER 10 ROTATION

Figure 10-11 A long rod is much easier to

rotate about (a) its central (longitudinal)

axis than about (b) an axis through its

center and perpendicular to its length. The

reason for the difference is that the mass

is distributed closer to the rotation axis in

(a) than in (b).

Rotation

axis

(a)

(b)

Rod is easy to rotate

this way.

Harder this way.

Instead, we shall treat the table saw (and any other rotating rigid body) as a

collection of particles with different speeds. We can then add up the kinetic

energies of all the particles to find the kinetic energy of the body as a whole.

In this way we obtain, for the kinetic energy of a rotating body,

(10-31)

in which miis the mass of the ith particle and viis its speed.The sum is taken over

all the particles in the body.

The problem with Eq. 10-31 is that viis not the same for all particles.We solve

this problem by substituting for vfrom Eq. 10-18 (vvr), so that we have

(10-32)

in which vis the same for all particles.

The quantity in parentheses on the right side of Eq. 10-32 tells us how

the mass of the rotating body is distributed about its axis of rotation. We call

that quantity the rotational inertia (or moment of inertia)Iof the body with

respect to the axis of rotation. It is a constant for a particular rigid body and

a particular rotation axis. (Caution: That axis must always be specified if the

value of Iis to be meaningful.)

We may now write

(rotational inertia) (10-33)

and substitute into Eq. 10-32, obtaining

(radian measure) (10-34)

as the expression we seek. Because we have used the relation vvrin deriving

Eq. 10-34, vmust be expressed in radian measure. The SI unit for Iis the

kilogram–square meter (kgm2).

The Plan. If we have a few particles and a specified rotation axis, we find mr2

for each particle and then add the results as in Eq. 10-33 to get the total rotational in-

ertia I. If we want the total rotational kinetic energy, we can then substitute that I

into Eq. 10-34.That is the plan for a few particles, but suppose we have a huge num-

ber of particles such as in a rod. In the next module we shall see how to handle such

continuous bodies and do the calculation in only a few minutes.

Equation 10-34, which gives the kinetic energy of a rigid body in pure rotation,

is the angular equivalent of the formula , which gives the kinetic energyK1

2Mvcom

K1



I



miri

K



2mi(vri)21





miri



v2,





2mivi

K1

2m1v2

11

2m2v2

21

2m3v2

3 

of a rigid body in pure translation. In both formulas there is a factor of . Where

mass Mappears in one equation, I(which involves both mass and its distribution)

appears in the other. Finally, each equation contains as a factor the square of a

speed—translational or rotational as appropriate. The kinetic energies of transla-

tion and of rotation are not different kinds of energy. They are both kinetic energy,

expressed in ways that are appropriate to the motion at hand.

We noted previously that the rotational inertia of a rotating body involves

not only its mass but also how that mass is distributed. Here is an example that

you can literally feel. Rotate a long, fairly heavy rod (a pole, a length of lumber,

or something similar), first around its central (longitudinal) axis (Fig. 10-11a)

and then around an axis perpendicular to the rod and through the center

(Fig. 10-11b). Both rotations involve the very same mass, but the first rotation is

much easier than the second. The reason is that the mass is distributed much

closer to the rotation axis in the first rotation. As a result, the rotational inertia

of the rod is much smaller in Fig. 10-11athan in Fig. 10-11b. In general, smaller

rotational inertia means easier rotation.

273

10-5 CALCULATING THE ROTATIONAL INERTIA

Checkpoint 4

The figure shows three small spheres that rotate

about a vertical axis.The perpendicular distance

between the axis and the center of each sphere is

given. Rank the three spheres according to their

rotational inertia about that axis,greatest first.

Rotation

axis

4 kg

3 m

2 m

1 m

9 kg

36 kg

10-5 CALCULATING THE ROTATIONAL INERTIA

After reading this module, you should be able to . . .

10.20 Determine the rotational inertia of a body if it is given in

Table 10-2.

10.21 Calculate the rotational inertia of a body by integration

over the mass elements of the body.

10.22 Apply the parallel-axis theorem for a rotation axis that is

displaced from a parallel axis through the center of mass of

a body.

●Iis the rotational inertia of the body, defined as

for a system of discrete particles and defined as

for a body with continuously distributed mass. The rand riin

these expressions represent the perpendicular distance from

the axis of rotation to each mass element in the body, and the

integration is carried out over the entire body so as to include

every mass element.

Ir2dm

I



miri

●The parallel-axis theorem relates the rotational inertia Iof a

body about any axis to that of the same body about a parallel

axis through the center of mass:

IIcom Mh2.

Here his the perpendicular distance between the two axes,

and Icom is the rotational inertia of the body about the axis

through the com. We can describe has being the distance

the actual rotation axis has been shifted from the rotation axis

through the com.

Learning Objectives

Key Ideas

Calculating the Rotational Inertia

If a rigid body consists of a few particles, we can calculate its rotational inertia

about a given rotation axis with Eq. 10-33 ; that is, we can find the

product mr2for each particle and then sum the products. (Recall that ris the per-

pendicular distance a particle is from the given rotation axis.)

If a rigid body consists of a great many adjacent particles (it is continuous, like

a Frisbee), using Eq. 10-33 would require a computer.Thus, instead, we replace the

sum in Eq. 10-33 with an integral and define the rotational inertia of the body as

(rotational inertia, continuous body). (10-35)

Table 10-2 gives the results of such integration for nine common body shapes and

the indicated axes of rotation.

Parallel-Axis Theorem

Suppose we want to find the rotational inertia Iof a body of mass Mabout a

given axis. In principle, we can always find Iwith the integration of Eq. 10-35.

However, there is a neat shortcut if we happen to already know the rotational in-

ertia Icom of the body about a parallel axis that extends through the body’s center

of mass. Let hbe the perpendicular distance between the given axis and the axis

Ir2dm

(Imiri

274 CHAPTER 10 ROTATION

Table 10-2 Some Rotational Inertias

Axis

Hoop about

central axis

Axis

Annular cylinder

(or ring) about

central axis

I=MR2(b)(a)I=M(R1

2+R2

Thin rod about

axis through center

perpendicular to

length

(e)

I=ML2

Axis

Hoop about any

diameter

Slab about

perpendicular

axis through

center

(i)(h)

I=MR2I= M(a2+b2)

Axis

Solid cylinder

(or disk) about

central axis

(c)

I=MR2

Axis

Solid cylinder

(or disk) about

central diameter

(d)

I=MR2+ML2

Axis

Thin

spherical shell

about any

diameter

(g)

I=MR2

Solid sphere

about any

diameter

(f)

I=MR2

Axis

Figure 10-12 A rigid body in cross section,

with its center of mass at O.The parallel-

axis theorem (Eq. 10-36) relates the

rotational inertia of the body about an axis

through Oto that about a parallel axis

through a point such as P, a distance h

from the body’s center of mass.

x– a

y– b

com

Rotation axis

through

center of mass

Rotation axis

through P

We need to relate the rotational inertia

around the axis at P to that around the

axis at the com.

through the center of mass (remember these two axes must be parallel).Then the

rotational inertia Iabout the given axis is

IIcom Mh2(parallel-axis theorem). (10-36)

Think of the distance has being the distance we have shifted the rotation axis

from being through the com.This equation is known as the parallel-axis theorem.

We shall now prove it.

Proof of the Parallel-Axis Theorem

Let Obe the center of mass of the arbitrarily shaped body shown in cross section

in Fig. 10-12. Place the origin of the coordinates at O. Consider an axis through O

perpendicular to the plane of the figure, and another axis through point Pparal-

lel to the first axis. Let the xand ycoordinates of Pbe aand b.

Let dm be a mass element with the general coordinates xand y. The rota-

tional inertia of the body about the axis through Pis then, from Eq. 10-35,

which we can rearrange as

(10-37)

From the definition of the center of mass (Eq. 9-9), the middle two integrals of

Eq. 10-37 give the coordinates of the center of mass (multiplied by a constant)

I (x2y2)dm 2axdm2bydm(a2b2)dm.

Ir2dm [(xa)2(yb)2]dm,

275

10-5 CALCULATING THE ROTATIONAL INERTIA

and thus must each be zero. Because x2y2is equal to R2, where Ris the dis-

tance from Oto dm, the first integral is simply Icom, the rotational inertia of the

body about an axis through its center of mass. Inspection of Fig. 10-12 shows that

the last term in Eq. 10-37 is Mh2, where Mis the body’s total mass. Thus,

Eq. 10-37 reduces to Eq. 10-36, which is the relation that we set out to prove.

Checkpoint 5

The figure shows a book-like object (one side is

longer than the other) and four choices of rotation

axes,all perpendicular to the face of the object.

Rank the choices according to the rotational inertia

of the object about the axis,greatest first.

(

)(

)

left and Lfor the particle on the right. Now Eq. 10-33

gives us

Im(0)2mL2mL2. (Answer)

Second technique: Because we already know Icom about an

axis through the center of mass and because the axis here is

parallel to that “com axis,” we can apply the parallel-axis

theorem (Eq. 10-36).We find

(Answer)mL2.

IIcom Mh21

2mL2(2m)(1

2L)2

Sample Problem 10.06 Rotational inertia of a two-particle system

Figure 10-13ashows a rigid body consisting of two particles of

mass mconnected by a rod of length Land negligible mass.

(a) What is the rotational inertia Icom about an axis through the

center of mass,perpendicular to the rod as shown?

KEY IDEA

Because we have only two particles with mass, we can find

the body’s rotational inertia Icom by using Eq. 10-33 rather

than by integration. That is, we find the rotational inertia of

each particle and then just add the results.

Calculations: For the two particles, each at perpendicular

distance from the rotation axis, we have

(Answer)

(b) What is the rotational inertia Iof the body about an axis

through the left end of the rod and parallel to the first axis

(Fig. 10-13b)?

KEY IDEAS

This situation is simple enough that we can find Iusing

either of two techniques. The first is similar to the one used

in part (a). The other, more powerful one is to apply the

parallel-axis theorem.

First technique: We calculate Ias in part (a), except here

the perpendicular distance riis zero for the particle on the

1

2mL2.

I



miri

2(m)(1

2L)2(m)(1

2L)2

Additional examples, video, and practice available at WileyPLUS

m m

(a)

com

Rotat

on ax

through

center of mass

(b)

com

Rotation axis through

end of rod

Here the rotation axis is through the com.

Here it has been shifted from the com

without changing the orientation. We

can use the parallel-axis theorem.

Figure 10-13 A rigid body consisting of two particles of mass m

joined by a rod of negligible mass.

276 CHAPTER 10 ROTATION

Sample Problem 10.07 Rotational inertia of a uniform rod, integration

Figure 10-14 shows a thin, uniform rod of mass Mand length

L, on an xaxis with the origin at the rod’s center.

(a) What is the rotational inertia of the rod about the

perpendicular rotation axis through the center?

KEY IDEAS

(1) The rod consists of a huge number of particles at a great

many different distances from the rotation axis. We certainly

don’t want to sum their rotational inertias individually. So, we

first write a general expression for the rotational inertia of a

mass element dm at distance rfrom the rotation axis: r2dm.

(2) Then we sum all such rotational inertias by integrating the

expression (rather than adding them up one by one). From

Eq. 10-35,we write

(10-38)

(3) Because the rod is uniform and the rotation axis is at the

center, we are actually calculating the rotational inertia Icom

about the center of mass.

Calculations: We want to integrate with respect to coordinate

x(not mass mas indicated in the integral), so we must relate

the mass dm of an element of the rod to its length dx along the

rod. (Such an element is shown in Fig. 10-14.) Because the rod

is uniform, the ratio of mass to length is the same for all the el-

ements and for the rod as a whole.Thus, we can write

or dm M

Ldx.

element’s mass dm

element’s length dx rod’s mass M

rod’s length L

Ir2dm.

Figure 10-14 A uniform rod of length L

and mass M. An element of mass dm

and length dx is represented.

We can now substitute this result for dm and xfor rin

Eq. 10-38.Then we integrate from end to end of the rod (from

xL/2 to xL/2) to include all the elements.We find

(Answer)

(b) What is the rod’s rotational inertia Iabout a new rotation

axis that is perpendicular to the rod and through the left end?

KEY IDEAS

We can find Iby shifting the origin of the xaxis to the left end

of the rod and then integrating from to .However,

here we shall use a more powerful (and easier) technique by

applying the parallel-axis theorem (Eq. 10-36), in which we

shift the rotation axis without changing its orientation.

Calculations: If we place the axis at the rod’s end so that it

is parallel to the axis through the center of mass, then we

can use the parallel-axis theorem (Eq. 10-36). We know

from part (a) that Icom is . From Fig. 10-14, the perpen-

dicular distance hbetween the new rotation axis and the

center of mass is . Equation 10-36 then gives us

(Answer)

Actually, this result holds for any axis through the left

or right end that is perpendicular to the rod.

1

3ML2.

IIcom Mh21

12 ML2(M)(1

2L)2

12 ML2

xLx 0

1

12 ML2.

M





L/2

L/2

M











L





IxL/2

xL/2 x2





Additional examples, video, and practice available at WileyPLUS

Rotation

axis

com M

We want the

rotational inertia.

Rotation

axis

xdm

First, pick any tiny element

and write its rotational

inertia as x2dm.

x=−

Rotation

axis

Leftmost Rightmost

2x=L

Then, using integration, add up

the rotational inertias for all of

the elements, from leftmost to

rightmost.

277

10-6 TORQUE

KEY IDEA

The released energy was equal to the rotational kinetic en-

ergy Kof the rotor just as it reached the angular speed of

14 000 rev/min.

Calculations: We can find Kwith Eq. 10-34 , but

first we need an expression for the rotational inertia I. Because

the rotor was a disk that rotated like a merry-go-round, Iis

given in Table 10-2c.Thus,

The angular speed of the rotor was

Then, with Eq. 10-34, we find the (huge) energy release:

(Answer)2.1 107 J.

K1

2Iv21

2(19.64 kgm2)(1.466 103 rad/s)2

1.466 103 rad/s.

v(14 000 rev/min)(2p rad/rev)



1 min

60 s



I1

2MR21

2(272 kg)(0.38 m)219.64 kgm2.

(I1

2MR2)

(K1

2Iv2)

Sample Problem 10.08 Rotational kinetic energy, spin test explosion

Large machine components that undergo prolonged, high-

speed rotation are first examined for the possibility of fail-

ure in a spin test system. In this system, a component is spun

up (brought up to high speed) while inside a cylindrical

arrangement of lead bricks and containment liner, all within

a steel shell that is closed by a lid clamped into place. If the

rotation causes the component to shatter, the soft lead

bricks are supposed to catch the pieces for later analysis.

In 1985, Test Devices, Inc. (www.testdevices.com) was spin

testing a sample of a solid steel rotor (a disk) of mass M

272 kg and radius R38.0 cm. When the sample reached

an angular speed vof 14 000 rev/min, the test engineers

heard a dull thump from the test system, which was

located one floor down and one room over from them.

Investigating, they found that lead bricks had been thrown

out in the hallway leading to the test room, a door to the

room had been hurled into the adjacent parking lot, one

lead brick had shot from the test site through the wall of a

neighbor’s kitchen, the structural beams of the test build-

ing had been damaged, the concrete floor beneath the

spin chamber had been shoved downward by about 0.5

cm, and the 900 kg lid had been blown upward through

the ceiling and had then crashed back onto the test equip-

ment (Fig. 10-15). The exploding pieces had not pene-

trated the room of the test engineers only by luck.

How much energy was released in the explosion of the

rotor?

Figure 10-15 Some of the

destruction caused by

the explosion of a rap-

idly rotating steel disk.

Courtesy Test Devices, Inc.

10-6 TORQUE

After reading this module, you should be able to . . .

10.23 Identify that a torque on a body involves a force and a

position vector, which extends from a rotation axis to the

point where the force is applied.

10.24 Calculate the torque by using (a) the angle between

the position vector and the force vector, (b) the line of ac-

tion and the moment arm of the force, and (c) the force

component perpendicular to the position vector.

10.25 Identify that a rotation axis must always be specified to

calculate a torque.

10.26 Identify that a torque is assigned a positive or negative

sign depending on the direction it tends to make the body

rotate about a specified rotation axis: “clocks are negative.”

10.27 When more than one torque acts on a body about a

rotation axis, calculate the net torque.

Learning Objectives

●Torque is a turning or twisting action on a body about a

rotation axis due to a force . If is exerted at a point given

by the position vector relative to the axis, then the magni-

tude of the torque is

where Ftis the component of perpendicular to and

fis the angle between and . The quantity is the r

trF

trFrF sin f,

:perpendicular distance between the rotation axis and

an extended line running through the vector. This line

is called the line of action of , and is called the

moment arm of . Similarly, ris the moment arm of Ft.

●The SI unit of torque is the newton-meter (Nm). A

torque tis positive if it tends to rotate a body at rest

counterclockwise and negative if it tends to rotate the

body clockwise.

:r

Key Ideas

Additional examples, video, and practice available at WileyPLUS

278 CHAPTER 10 ROTATION

Checkpoint 6

The figure shows an overhead view of a meter stick that can pivot about the dot at the position

marked 20 (for 20 cm).All five forces on the stick are horizontal and have the same magnitude.

Rank the forces according to the magnitude of the torque they produce, greatest first.

0 20 40

vot po

100

F2F3

Figure 10-16 (a) A force acts on a rigid

body, with a rotation axis perpendicular to

the page. The torque can be found with

(a) angle f,(b) tangential force compo-

nent Ft, or (c) moment arm .r

(a)

(b)

(c)

FrFt

Rotation

axis

Rotation

axis

Line of

action of F

Moment arm

of F

Rotation

axis

The torque due to this force

causes rotation around this axis

(which extends out toward you).

You calculate the same torque by

using this moment arm distance

and the full force magnitude.

But actually only the tangential

component of the force causes

the rotation.

magnitude FtFsin f.This component does cause rotation.

Calculating Torques. The ability of to rotate the body depends not only

on the magnitude of its tangential component Ft, but also on just how far from O

the force is applied. To include both these factors, we define a quantity called

torque tas the product of the two factors and write it as

t(r)(Fsin f). (10-39)

Two equivalent ways of computing the torque are

t(r)(Fsin f)rFt(10-40)

and (10-41)

where is the perpendicular distance between the rotation axis at Oand an extendedr

t(r sin f)(F)rF,

Torque

A doorknob is located as far as possible from the door’s hinge line for a good rea-

son. If you want to open a heavy door, you must certainly apply a force, but that

is not enough.Where you apply that force and in what direction you push are also

important. If you apply your force nearer to the hinge line than the knob, or at

any angle other than 90to the plane of the door, you must use a greater force

than if you apply the force at the knob and perpendicular to the door’s plane.

Figure 10-16ashows a cross section of a body that is free to rotate about an

axis passing through Oand perpendicular to the cross section. A force is

applied at point P, whose position relative to Ois defined by a position vector .

The directions of vectors and make an angle fwith each other. (For simplic-

ity, we consider only forces that have no component parallel to the rotation axis;

thus, is in the plane of the page.)

To determine how results in a rotation of the body around the rotation

axis, we resolve into two components (Fig. 10-16b). One component, called the

radial component Fr, points along . This component does not cause rotation,

because it acts along a line that extends through O. (If you pull on a door par-

allel to the plane of the door, you do not rotate the door.) The other compo-

nent of , called the tangential component Ft, is perpendicular to and hasr

line running through the vector (Fig. 10-16c). This extended line is called the line

of action of , and is called the moment arm of . Figure 10-16bshows that we

can describe r,the magnitude of ,as being the moment arm of the force component Ft.

Torque, which comes from the Latin word meaning “to twist,” may be loosely

identified as the turning or twisting action of the force .When you apply a force

to an object—such as a screwdriver or torque wrench—with the purpose of turn-

ing that object, you are applying a torque. The SI unit of torque is the newton-

meter (Nm). Caution: The newton-meter is also the unit of work. Torque and

work, however, are quite different quantities and must not be confused. Work is

often expressed in joules (1 J 1Nm), but torque never is.

Clocks Are Negative. In Chapter 11 we shall use vector notation for torques,

but here, with rotation around a single axis, we use only an algebraic sign. If a

torque would cause counterclockwise rotation, it is positive. If it would cause

clockwise rotation, it is negative. (The phrase “clocks are negative” from Module

10-1 still works.)

Torques obey the superposition principle that we discussed in Chapter 5 for

forces:When several torques act on a body, the net torque (or resultant torque) is

the sum of the individual torques.The symbol for net torque is tnet.

r

279

10-7 NEWTON’S SECOND LAW FOR ROTATION

After reading this module, you should be able to . . .

10.28 Apply Newton’s second law for rotation to relate the

net torque on a body to the body’s rotational inertia and

rotational acceleration, all calculated relative to a specified

rotation axis.

●The rotational analog of Newton’s second law is

tnet Ia,

where tnet is the net torque acting on a particle or rigid body,

Iis the rotational inertia of the particle or body about the

rotation axis, and ais the resulting angular acceleration about

that axis.

Learning Objective

Key Idea

Newton’s Second Law for Rotation

A torque can cause rotation of a rigid body, as when you use a torque to rotate

a door. Here we want to relate the net torque tnet on a rigid body to the angular

acceleration athat torque causes about a rotation axis. We do so by analogy with

Newton’s second law (Fnet ma) for the acceleration aof a body of mass mdue

to a net force Fnet along a coordinate axis.We replace Fnet with tnet,mwith I, and a

with ain radian measure, writing

tnet Ia(Newton’s second law for rotation). (10-42)

Proof of Equation 10-42

We prove Eq. 10-42 by first considering the simple situation shown in Fig. 10-17.

The rigid body there consists of a particle of mass mon one end of a massless rod

of length r.The rod can move only by rotating about its other end, around a rota-

tion axis (an axle) that is perpendicular to the plane of the page.Thus, the particle

can move only in a circular path that has the rotation axis at its center.

A force acts on the particle. However, because the particle can move

only along the circular path, only the tangential component Ftof the force (the

component that is tangent to the circular path) can accelerate the particle along

the path. We can relate Ftto the particle’s tangential acceleration atalong the

path with Newton’s second law, writing

Ftmat.

The torque acting on the particle is, from Eq. 10-40,

tFtrmatr.

From Eq. 10-22 (atar) we can write this as

tm(ar)r(mr2)a. (10-43)

The quantity in parentheses on the right is the rotational inertia of the particle

about the rotation axis (see Eq. 10-33, but here we have only a single particle).

Thus, using Ifor the rotational inertia, Eq. 10-43 reduces to

tIa(radian measure). (10-44)

If more than one force is applied to the particle, Eq. 10-44 becomes

tnet Ia(radian measure), (10-45)

which we set out to prove.We can extend this equation to any rigid body rotating

about a fixed axis, because any such body can always be analyzed as an assembly

of single particles.

Figure 10-17 A simple rigid body, free to

rotate about an axis through O, consists of

a particle of mass mfastened to the end of

a rod of length rand negligible mass. An

applied force causes the body to rotate.F

Rod

Rotation axis

The torque due to the tangential

component of the force causes

an angular acceleration around

the rotation axis.

280 CHAPTER 10 ROTATION

Additional examples, video, and practice available at WileyPLUS

KEY IDEA

Because the moment arm for is no longer zero, the torqueF

Checkpoint 7

The figure shows an overhead view of a meter stick that can pivot about the point indicated, which is

to the left of the stick’s midpoint.Two horizontal forces, and ,are applied to the stick.Only is

shown. Force is perpendicular to the stick and is applied at the right end. If the stick is not to turn,

(a) what should be the direction of , and (b) should F2be greater than,less than, or equal to F1?F

Pivot point

Sample Problem 10.09 Using Newton’s second law for rotation in a basic judo hip throw

To throw an 80 kg opponent with a basic judo hip throw, you

intend to pull his uniform with a force and a moment arm

d10.30 m from a pivot point (rotation axis) on your right

hip (Fig. 10-18). You wish to rotate him about the pivot

point with an angular acceleration aof 6.0 rad/s2—that is,

with an angular acceleration that is clockwise in the figure.

Assume that his rotational inertia Irelative to the pivot

point is 15 kgm2.

(a) What must the magnitude of be if, before you throw

him, you bend your opponent forward to bring his center of

mass to your hip (Fig.10-18a)?

KEY IDEA

We can relate your pull on your opponent to the given an-

gular acceleration avia Newton’s second law for rotation

(tnet Ia).

Calculations: As his feet leave the floor, we can assume that

only three forces act on him: your pull , a force on him

from you at the pivot point (this force is not indicated in Fig.

10-18), and the gravitational force .To use tnet Ia, we need

the corresponding three torques, each about the pivot point.

From Eq. 10-41 (tF), the torque due to your pull F

r

Figure 10-18 A judo hip throw (a) correctly executed and (b) incor-

rectly executed.

Opponent's

center of

mass

Moment arm d1

of your pull

Pivot

on hip

Moment arm d2

of gravitational

force on

opponent

Moment

arm d1

of your pull

(a) (b)

is equal to F, where is the moment arm and the

sign indicates the clockwise rotation this torque tends to

cause. The torque due to is zero, because acts at theN

r

pivot point and thus has moment arm 0.

To evaluate the torque due to , we can assume that

acts at your opponent’s center of mass. With the center of

mass at the pivot point, has moment arm 0 and thusr

r

ponent is due to your pull , and we can write tnet Iaas

d1FIa.

We then find

300 N. (Answer)

(b) What must the magnitude of be if your opponent

remains upright before you throw him, so that has a mo-

ment arm d20.12 m (Fig.10-18b)?

FIa

(15 kgm2)(6.0 rad/s2)

0.30 m

the torque due to is zero. So, the only torque on your op-

due to is now equal to d2mg and is positive because the

torque attempts counterclockwise rotation.

Calculations: Now we write tnet Iaas

d1Fd2mg Ia,

which gives

From (a), we know that the first term on the right is equal to

300 N. Substituting this and the given data, we have

613.6 N 610 N. (Answer)

The results indicate that you will have to pull much harder if

you do not initially bend your opponent to bring his center

of mass to your hip. A good judo fighter knows this lesson

from physics. Indeed, physics is the basis of most of the mar-

tial arts, figured out after countless hours of trial and error

over the centuries.



F300 N (0.12 m)(80 kg)(9.8 m/s2)

0.30 m

FIa

d2mg

281

10-7 NEWTON’S SECOND LAW FOR ROTATION

Sample Problem 10.10 Newton’s second law, rotation, torque, disk

Figure 10-19ashows a uniform disk, with mass M2.5 kg

and radius R20 cm, mounted on a fixed horizontal axle.

A block with mass m1.2 kg hangs from a massless cord that

is wrapped around the rim of the disk. Find the acceleration of

the falling block, the angular acceleration of the disk, and the

tension in the cord.The cord does not slip, and there is no fric-

tion at the axle.

KEY IDEAS

(1) Taking the block as a system, we can relate its accelera-

tion ato the forces acting on it with Newton’s second law

( ). (2) Taking the disk as a system, we can relate

its angular acceleration ato the torque acting on it with

Newton’s second law for rotation (tnet Ia). (3) To combine

the motions of block and disk, we use the fact that the linear

acceleration aof the block and the (tangential) linear accel-

eration of the disk rim are equal. (To avoid confusion

about signs, let’s work with acceleration magnitudes and

explicit algebraic signs.)

Forces on block: The forces are shown in the block’s free-

body diagram in Fig. 10-19b: The force from the cord is ,

and the gravitational force is , of magnitude mg. We can

now write Newton’s second law for components along a ver-

tical yaxis (Fnet,ymay) as

Tmg m(a), (10-46)

where ais the magnitude of the acceleration (down the y

axis). However, we cannot solve this equation for abecause

it also contains the unknown T.

Torque on disk: Previously, when we got stuck on the yaxis,

we switched to the xaxis. Here, we switch to the rotation of

the disk and use Newton’s second law in angular form. To

calculate the torques and the rotational inertia I, we take

the rotation axis to be perpendicular to the disk and through

its center,at point Oin Fig. 10-19c.

The torques are then given by Eq. 10-40 (trFt). The

gravitational force on the disk and the force on the disk from

the axle both act at the center of the disk and thus at distance

r0, so their torques are zero.The force on the disk due to

the cord acts at distance rRand is tangent to the rim of the

disk. Therefore, its torque is RT, negative because the

torque rotates the disk clockwise from rest. Let abe the mag-

nitude of the negative (clockwise) angular acceleration. From

Table 10-2c, the rotational inertia Iof the disk is . Thus

we can write the general equation tnet Iaas

(10-47)RT 1

2MR2(a).

2MR2

net m:

This equation seems useless because it has two

unknowns, aand T, neither of which is the desired a.

However, mustering physics courage, we can make it useful

with this fact: Because the cord does not slip, the magnitude

aof the block’s linear acceleration and the magnitude atof

the (tangential) linear acceleration of the rim of the disk are

equal. Then, by Eq. 10-22 (atar) we see that here a

a/R. Substituting this in Eq. 10-47 yields

(10-48)

Combining results: Combining Eqs. 10-46 and 10-48 leads to

. (Answer)

We then use Eq. 10-48 to find T:

(Answer)

As we should expect, acceleration aof the falling block is less

than g, and tension Tin the cord (6.0 N) is less than the

gravitational force on the hanging block (mg 11.8 N).

We see also that aand Tdepend on the mass of the disk but

not on its radius.

As a check, we note that the formulas derived above

predict agand T0 for the case of a massless disk (M

0). This is what we would expect; the block simply falls as a

free body. From Eq. 10-22, the magnitude of the angular ac-

celeration of the disk is

(Answer)aa

R4.8 m/s2

0.20 m 24 rad/s2.

6.0 N.

T1

2Ma 1

2(2.5 kg)(4.8 m/s2)

4.8 m/s2

ag2m

M2m(9.8 m/s2)(2)(1.2 kg)

2.5 kg (2)(1.2 kg)

T1

2Ma.

(b)(a)

(c)

The torque due to the

cord's pull on the rim

causes an angular

acceleration of the disk.

These two forces

determine the block's

(linear) acceleration.

We need to relate

those two

accelerations.

Figure 10-19 (a) The falling block causes the disk to rotate. (b) A

free-body diagram for the block. (c) An incomplete free-body

diagram for the disk.

Additional examples, video, and practice available at WileyPLUS

282 CHAPTER 10 ROTATION

Work and Rotational Kinetic Energy

As we discussed in Chapter 7, when a force Fcauses a rigid body of mass mto ac-

celerate along a coordinate axis, the force does work Won the body. Thus, the

body’s kinetic energy can change. Suppose it is the only energy of the(K1

2mv2)

10-8 WORK AND ROTATIONAL KINETIC ENERGY

After reading this module, you should be able to . . .

10.29 Calculate the work done by a torque acting on a rotat-

ing body by integrating the torque with respect to the an-

gle of rotation.

10.30 Apply the work–kinetic energy theorem to relate the

work done by a torque to the resulting change in the rota-

tional kinetic energy of the body.

10.31 Calculate the work done by a constant torque by relat-

ing the work to the angle through which the body rotates.

10.32 Calculate the power of a torque by finding the rate at

which work is done.

10.33 Calculate the power of a torque at any given instant by

relating it to the torque and the angular velocity at that instant.

●The equations used for calculating work and power in rota-

tional motion correspond to equations used for translational

motion and are

and PdW

dt tv.

Wu

●When tis constant, the integral reduces to

Wt(ufui).

●The form of the work– kinetic energy theorem used for

rotating bodies is

KKfKi1

2Ivf

21



2W.

Learning Objectives

Key Ideas

body that changes.Then we relate the change Kin kinetic energy to the work W

with the work–kinetic energy theorem (Eq. 7-10), writing

(work–kinetic energy theorem). (10-49)

For motion confined to an xaxis, we can calculate the work with Eq. 7-32,

(work, one-dimensional motion). (10-50)

This reduces to WFd when Fis constant and the body’s displacement is d.

The rate at which the work is done is the power, which we can find with Eqs. 7-43

and 7-48,

(power,one-dimensional motion). (10-51)

Now let us consider a rotational situation that is similar. When a torque

accelerates a rigid body in rotation about a fixed axis, the torque does work W

on the body. Therefore, the body’s rotational kinetic energy can

change. Suppose that it is the only energy of the body that changes. Then we

can still relate the change Kin kinetic energy to the work Wwith the

work–kinetic energy theorem, except now the kinetic energy is a rotational

kinetic energy:

(work–kinetic energy theorem). (10-52)

Here, Iis the rotational inertia of the body about the fixed axis and viand vfare

the angular speeds of the body before and after the work is done.

KKfKi1

2Ivf

21



2W

(K1



PdW

dt Fv

Wxf

Fdx

KKfKi1

2mvf

21

2mvi

2W

283

10-8 WORK AND ROTATIONAL KINETIC ENERGY

Also, we can calculate the work with a rotational equivalent of Eq. 10-50,

(work, rotation about fixed axis), (10-53)

where tis the torque doing the work W, and uiand ufare the body’s angular

positions before and after the work is done, respectively. When tis constant,

Eq. 10-53 reduces to

Wt(ufui)(work, constant torque). (10-54)

The rate at which the work is done is the power, which we can find with the rota-

tional equivalent of Eq. 10-51,

(power,rotation about fixed axis). (10-55)

Table 10-3 summarizes the equations that apply to the rotation of a rigid body

about a fixed axis and the corresponding equations for translational motion.

Proof of Eqs. 10-52 through 10-55

Let us again consider the situation of Fig. 10-17, in which force rotates a rigid

body consisting of a single particle of mass mfastened to the end of a massless

rod. During the rotation, force does work on the body. Let us assume that the

only energy of the body that is changed by is the kinetic energy. Then we can

apply the work–kinetic energy theorem of Eq. 10-49:

KKfKiW. (10-56)

Using and Eq. 10-18 (vvr), we can rewrite Eq. 10-56 as

(10-57)

From Eq. 10-33, the rotational inertia for this one-particle body is Imr2.

Substituting this into Eq. 10-57 yields

which is Eq. 10-52.We derived it for a rigid body with one particle, but it holds for

any rigid body rotated about a fixed axis.

We next relate the work Wdone on the body in Fig. 10-17 to the torque t

on the body due to force . When the particle moves a distance ds along itsF

K1

2Ivf

21



2W,

K1

2mr2vf

21

2mr2vi

2W.

K1

2mv2

PdW

dt tv

Wuf

tdu

Table 10-3 Some Corresponding Relations for Translational and Rotational Motion

Pure Translation (Fixed Direction) Pure Rotation (Fixed Axis)

Position xAngular position u

Velocity vdx/dt Angular velocity vdu/dt

Acceleration adv/dt Angular acceleration adv/dt

Mass mRotational inertia I

Newton’s second law Fnet ma Newton’s second law tnet Ia

Work WF dx Work Wtdu

Kinetic energy Kinetic energy K1

2Iv2

K1

2mv2

Power (constant force) PFv Power (constant torque) Ptv

Work–kinetic energy theorem WKWork–kinetic energy theorem WK

284 CHAPTER 10 ROTATION

Sample Problem 10.11 Work, rotational kinetic energy, torque, disk

Let the disk in Fig. 10-19 start from rest at time t0 and

also let the tension in the massless cord be 6.0 N and the an-

gular acceleration of the disk be 24 rad/s2.What is its rota-

tional kinetic energy Kat t2.5 s?

KEY IDEA

We can find Kwith Eq. 10-34 We already know(K1

2Iv2).

Calculations: First, we relate the change in the kinetic

energy of the disk to the net work Wdone on the disk, using

the work–kinetic energy theorem of Eq. 10-52 (KfKiW).

With Ksubstituted for Kfand 0 for Ki,we get

KKiW0WW. (10-60)

Next we want to find the work W. We can relate Wto

the torques acting on the disk with Eq. 10-53 or 10-54. The

only torque causing angular acceleration and doing work is

the torque due to force on the disk from the cord, which isT

that , but we do not yet know vat t2.5 s.

However, because the angular acceleration ahas the con-

stant value of 24 rad/s2, we can apply the equations for

constant angular acceleration in Table 10-1.

Calculations: Because we want vand know aand v0(0),

we use Eq. 10-12:

vv0at0atat.

Substituting vatand into Eq.10-34, we find

(Answer)

KEY IDEA

We can also get this answer by finding the disk’s kinetic

energy from the work done on the disk.

90 J.

1

4(2.5 kg)[(0.20 m)(24 rad/s2)(2.5 s)]2

K1

2Iv21

2(1

2MR2)(at)21

4M(Rat)2

I1

2MR2

I1

2MR2

Additional examples, video, and practice available at WileyPLUS

circular path, only the tangential component Ftof the force accelerates the parti-

cle along the path. Therefore, only Ftdoes work on the particle. We write that

work dW as Ftds. However, we can replace ds with r du, where duis the angle

through which the particle moves.Thus we have

dW Ftr du. (10-58)

From Eq. 10-40, we see that the product Ftris equal to the torque t, so we can

rewrite Eq. 10-58 as

dW tdu. (10-59)

The work done during a finite angular displacement from uito ufis then

which is Eq. 10-53. It holds for any rigid body rotating about a fixed axis.

Equation 10-54 comes directly from Eq. 10-53.

We can find the power Pfor rotational motion from Eq. 10-59:

which is Eq. 10-55.

PdW

dt t

dt tv,

Wuf

tdu,

equal to TR. Because ais constant, this torque also must

be constant.Thus, we can use Eq. 10-54 to write

Wt(ufui)TR(ufui). (10-61)

Because ais constant, we can use Eq. 10-13 to find

ufui.With vi0, we have

Now we substitute this into Eq. 10-61 and then substitute the

result into Eq. 10-60. Inserting the given values T6.0 N

and a24 rad/s2, we have

(Answer)90 J.



2(6.0 N)(0.20 m)(24 rad/s2)(2.5 s)2

KWTR(ufui)TR(1

2at2)

2TRat2

ufuivit1

2at201

2at21

2at2

285

REVIEW & SUMMARY

Angular Position To describe the rotation of a rigid body about

a fixed axis, called the rotation axis, we assume a reference line is

fixed in the body, perpendicular to that axis and rotating with the

body.We measure the angular position uof this line relative to a fixed

direction.When uis measured in radians,

(radian measure), (10-1)

where sis the arc length of a circular path of radius rand angle u.

Radian measure is related to angle measure in revolutions and de-

grees by 1 rev 3602prad. (10-2)

Angular Displacement A body that rotates about a rotation

axis,changing its angular position from u1to u2, undergoes an angu-

lar displacement uu2u1, (10-4)

where uis positive for counterclockwise rotation and negative for

clockwise rotation.

Angular Velocity and Speed If a body rotates through an

angular displacement uin a time interval t, its average angular

velocity vavg is

(10-5)

The (instantaneous) angular velocity vof the body is

(10-6)

Both vavg and vare vectors, with directions given by the right-hand

rule of Fig. 10-6. They are positive for counterclockwise rotation

and negative for clockwise rotation. The magnitude of the body’s

angular velocity is the angular speed.

Angular Acceleration If the angular velocity of a body

changes from v1to v2in a time interval tt2t1, the average

angular acceleration aavg of the body is

(10-7)

The (instantaneous) angular acceleration aof the body is

(10-8)

Both aavg and aare vectors.

The Kinematic Equations for Constant Angular Accel-

eration Constant angular acceleration (aconstant) is an im-

portant special case of rotational motion. The appropriate kine-

matic equations,given in Table 10-1, are

vv0at, (10-12)

(10-13)

(10-14)

(10-15)

(10-16)

Linear and Angular Variables Related A point in a rigid

rotating body, at a perpendicular distance r from the rotation axis,

uu0vt1

2at2.

uu01

2(v0v)t,

v2v0

22a(uu0),

uu0v0t1

2at2,

adv

dt .

aavg v2v1

t2t1

v

t.

vdu

dt .

vavg u

t.

us

Review & Summary

moves in a circle with radius r. If the body rotates through an angle u,

the point moves along an arc with length sgiven by

sur(radian measure), (10-17)

where uis in radians.

The linear velocity of the point is tangent to the circle; the

point’s linear speed vis given by

vvr(radian measure), (10-18)

where vis the angular speed (in radians per second) of the body.

The linear acceleration of the point has both tangential and

radial components.The tangential component is

atar(radian measure), (10-22)

where ais the magnitude of the angular acceleration (in radians

per second-squared) of the body.The radial component of is

(radian measure). (10-23)

If the point moves in uniform circular motion, the period Tof

the motion for the point and the body is

(radian measure). (10-19, 10-20)

Rotational Kinetic Energy and Rotational Inertia The ki-

netic energy Kof a rigid body rotating about a fixed axis is given by

(radian measure), (10-34)

in which Iis the rotational inertia of the body, defined as

(10-33)

for a system of discrete particles and defined as

(10-35)

for a body with continuously distributed mass. The rand riin these

expressions represent the perpendicular distance from the axis of

rotation to each mass element in the body, and the integration is car-

ried out over the entire body so as to include every mass element.

The Parallel-Axis Theorem The parallel-axis theorem relates

the rotational inertia Iof a body about any axis to that of the same

body about a parallel axis through the center of mass:

IIcom Mh2. (10-36)

Here his the perpendicular distance between the two axes, and

Icom is the rotational inertia of the body about the axis through the

com. We can describe has being the distance the actual rotation

axis has been shifted from the rotation axis through the com.

Torque Torque is a turning or twisting action on a body about a ro-

tation axis due to a force . If is exerted at a point given by the po-

sition vector relative to the axis, then the magnitude of the torque is

(10-40, 10-41, 10-39)

where Ftis the component of perpendicular to and fis the an-

gle between and .The quantity is the perpendicular distance

between the rotation axis and an extended line running through

the vector. This line is called the line of action of , and is

called the moment arm of . Similarly, ris the moment arm of Ft.F

:r

r

trF

trFrF sin f,

Ir2dm

I



miri

K1

2Iv2

T2pr

v2p

arv2

rv2r

8Figure 10-25bshows an overhead view of a horizontal bar that

is rotated about the pivot point by two horizontal forces, and ,

with at angle fto the bar. Rank the following values of faccord-

ing to the magnitude of the angular acceleration of the bar, greatest

first: 90,70, and 110.

9Figure 10-26 shows a uniform metal plate

that had been square before 25% of it was

snipped off. Three lettered points are indicated.

Rank them according to the rotational inertia of

the plate around a perpendicular axis through

them, greatest first.

286 CHAPTER 10 ROTATION

angles during the rotation, which is

counterclockwise and at a constant

rate. However, we are to decrease the

angle uof without changing the

magnitude of . (a) To keep the an-

gular speed constant, should we in-

crease, decrease, or maintain the mag-

nitude of ? Do forces (b) and (c)

tend to rotate the disk clockwise or

counterclockwise?

6In the overhead view of Fig. 10-24,

five forces of the same magnitude act

on a strange merry-go-round; it is a

square that can rotate about point P,at

midlength along one of the edges.

Rank the forces according to the mag-

nitude of the torque they create about

point P,greatest first.

7Figure 10-25ais an overhead view

of a horizontal bar that can pivot; two horizontal forces act on the

bar, but it is stationary. If the angle between the bar and is nowF

1Figure 10-20 is a graph of the an-

gular velocity versus time for a disk

rotating like a merry-go-round. For a

point on the disk rim, rank the in-

stants a,b,c, and daccording to the

magnitude of the (a) tangential and

(b) radial acceleration, greatest first.

2Figure 10-21 shows plots of angu-

lar position uversus time tfor three

cases in which a disk is rotated like a

merry-go-round. In each case, the ro-

tation direction changes at a certain

angular position uchange. (a) For each

case, determine whether uchange is

clockwise or counterclockwise from

u0, or whether it is at u0. For

each case, determine (b) whether

vis zero before, after, or at t0

and (c) whether ais positive,negative,or zero.

3A force is applied to the rim of a disk that can rotate like

a merry-go-round, so as to change its angular velocity. Its initial

and final angular velocities, respectively, for four situations are:

(a) 2 rad/s, 5 rad/s; (b) 2 rad/s, 5 rad/s; (c) 2 rad/s, 5 rad/s; and

(d) 2 rad/s, 5 rad/s. Rank the situations according to the work

done by the torque due to the force, greatest first.

4Figure 10-22bis a graph of the angular position of the rotating

disk of Fig. 10-22a. Is the angular velocity of the disk positive,nega-

tive, or zero at (a) t1 s, (b) t2 s, and (c) t3 s? (d) Is the an-

gular acceleration positive or negative?

Questions

abcd

Figure 10-20 Question 1.

–90°

90°

Figure 10-21 Question 2.

Rotation axis

t(s)

(rad)

1 2 3

(a) (b)

Figure 10-22 Question 4.

Figure 10-23 Question 5.

Figure 10-24 Question 6.

F1P

Pivot point

F1F2Pivot point

(

)(

)

Figure 10-26

Question 9.

Figure 10-25 Questions 7 and 8.

The SI unit of torque is the newton-meter (Nm). A torque t

is positive if it tends to rotate a body at rest counterclockwise and

negative if it tends to rotate the body clockwise.

Newton’s Second Law in Angular Form The rotational

analog of Newton’s second law is

tnet Ia, (10-45)

where tnet is the net torque acting on a particle or rigid body, Iis the ro-

tational inertia of the particle or body about the rotation axis, and ais

the resulting angular acceleration about that axis.

Work and Rotational Kinetic Energy The equations used

for calculating work and power in rotational motion correspond to

equations used for translational motion and are

(10-53)

and (10-55)

When tis constant,Eq. 10-53 reduces to

Wt(ufui). (10-54)

The form of the work– kinetic energy theorem used for rotating

bodies is

(10-52)KKfKi1

2Ivf

21



2W.

PdW

dt tv.

Wu

5In Fig.10-23, two forces and act on a disk that turns about

its center like a merry-go-round.The forces maintain the indicated

decreased from 90and the bar is still not to turn, should F2be

made larger,made smaller,or left the same?

287

PROBLEMS

10 Figure 10-27 shows three flat disks (of the same radius) that

can rotate about their centers like merry-go-rounds. Each disk con-

sists of the same two materials, one denser than the other (density is

mass per unit volume). In disks 1 and 3, the denser material forms

the outer half of the disk area. In disk 2, it forms the inner half of the

disk area. Forces with identical magnitudes are applied tangentially

to the disk, either at the outer edge or at the interface of the two ma-

terials, as shown. Rank the disks according to (a) the torque about

the disk center, (b) the rotational inertia about the disk center, and

Denser

Disk 1

Denser

Disk 3

F F

Lighter

Disk 2

Figure 10-27 Question 10.

11 Figure 10-28ashows a meter stick, half wood and half steel,

that is pivoted at the wood end at O.A force is applied to the steel

end at a. In Fig. 10-28b, the stick is reversed and pivoted at the steel

end at O, and the same force is applied at the wood end at a. Is the

resulting angular acceleration of Fig. 10-28agreater than, less than, or

the same as that of Fig.10-28b?

1 m

26 kg

(a)

2 m

7 kg

(b)

3 m

3 kg

(c)

Module 10-1 Rotational Variables

•1 A good baseball pitcher can throw a baseball toward home

plate at 85 mi/h with a spin of 1800 rev/min. How many revolutions

does the baseball make on its way to home plate? For simplicity,

assume that the 60 ft path is a straight line.

•2 What is the angular speed of (a) the second hand, (b) the

minute hand, and (c) the hour hand of a smoothly running analog

watch? Answer in radians per second.

••3 When a slice of buttered toast is accidentally pushed

over the edge of a counter, it rotates as it falls. If the distance to the

floor is 76 cm and for rotation less than 1 rev, what are the (a)

smallest and (b) largest angular speeds that cause the toast to hit

and then topple to be butter-side down?

••4 The angular position of a point on a rotating wheel is given

by u2.0 4.0t22.0t3, where uis in radians and tis in seconds.At

t0, what are (a) the point’s angular position and (b) its angular ve-

locity? (c) What is its angular velocity at t4.0 s? (d) Calculate its an-

gular acceleration at t2.0 s. (e) Is its angular acceleration constant?

••5 A diver makes 2.5 revolutions on the way from a 10-m-high

platform to the water. Assuming zero initial vertical velocity, find

the average angular velocity during the dive.

••6 The angular position of a point on the rim of a rotating wheel is

given by u4.0t3.0t2t3, where uis in radians and tis in seconds.

What are the angular velocities at (a) t2.0 s and (b) t4.0 s?

that begins at t2.0 s and ends at t4.0 s? What are the instanta-

neous angular accelerations at (d) the beginning and (e) the end of

this time interval?

•••7 The wheel in Fig. 10-30 has eight equally spaced spokes and

a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5

rev/s.You want to shoot a 20-cm-long arrow parallel to this axle and

ILW

through the wheel without hitting any

of the spokes. Assume that the arrow

and the spokes are very thin. (a) What

minimum speed must the arrow have?

(b) Does it matter where between the

axle and rim of the wheel you aim? If

so,what is the best location?

•••8 The angular acceleration of a

wheel is a6.0t44.0t2, with ain ra-

Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

•–••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday

Problems

Figure 10-30 Problem 7.

(b)(a)

OO´a´

Figure 10-28

Question 11.

Figure 10-29 Question 12.

12 Figure 10-29 shows three disks,

each with a uniform distribution of

mass. The radii Rand masses Mare

indicated. Each disk can rotate

around its central axis (perpendicular

to the disk face and through the cen-

ter). Rank the disks according to their

rotational inertias calculated about

their central axes,greatest first.

dians per second-squared and tin seconds.At time t0, the wheel

has an angular velocity of 2.0 rad/s and an angular position of

1.0 rad.Write expressions for (a) the angular velocity (rad/s) and

(b) the angular position (rad) as functions of time (s).

Module 10-2 Rotation with Constant Angular Acceleration

•9 A drum rotates around its central axis at an angular velocity

of 12.60 rad/s. If the drum then slows at a constant rate of 4.20

rad/s2, (a) how much time does it take and (b) through what angle

does it rotate in coming to rest?

•10 Starting from rest, a disk rotates about its central axis with

constant angular acceleration. In 5.0 s, it rotates 25 rad. During that

time, what are the magnitudes of (a) the angular acceleration and

(b) the average angular velocity? (c) What is the instantaneous an-

gular velocity of the disk at the end of the 5.0 s? (d) With the angu-

lar acceleration unchanged, through what additional angle will the

disk turn during the next 5.0 s?

•11 A disk, initially rotating at 120 rad/s, is slowed down

with a constant angular acceleration of magnitude 4.0 rad/s2. (a) How

much time does the disk take to stop? (b) Through what angle does

the disk rotate during that time?

•12 The angular speed of an automobile engine is increased at a

constant rate from 1200 rev/min to 3000 rev/min in 12 s. (a) What is

a point on Earth’s surface at latitude 40N? (Earth rotates about

that axis.) (b) What is the linear speed vof the point? What are

••26 The flywheel of a steam engine runs with a constant angular ve-

locity of 150 rev/min.When steam is shut off,the friction of the bearings

and of the air stops the wheel in 2.2 h. (a) What is the constant angular

acceleration, in revolutions per minute-squared, of the wheel during

the slowdown? (b) How many revolutions does the wheel make before

stopping? (c) At the instant the flywheel is turning at 75 rev/min, what

is the tangential component of the linear acceleration of a flywheel par-

ticle that is 50 cm from the axis of rotation? (d) What is the magnitude

of the net linear acceleration of the particle in (c)?

••27 A seed is on a turntable rotating at rev/min, 6.0 cm

from the rotation axis.What are (a) the seed’s acceleration and (b)

the least coefficient of static friction to avoid slippage? (c) If the

turntable had undergone constant angular acceleration from rest

in 0.25 s,what is the least coefficient to avoid slippage?

••28 In Fig. 10-31, wheel Aof radius

rA10 cm is coupled by belt Bto

wheel Cof radius rC25 cm.The an-

gular speed of wheel Ais increased

from rest at a constant rate of

1.6 rad/s2. Find the time needed for

wheel Cto reach an angular speed of

100 rev/min, assuming the belt does

not slip. (Hint: If the belt does not slip, the linear speeds at the two

rims must be equal.)

••29 Figure 10-32 shows an early method of measuring the

speed of light that makes use of a rotating slotted wheel.A beam of

331

celeration (in revolutions per minute-squared) will increase the

wheel’s angular speed to 1000 rev/min in 60.0 s? (d) How many

revolutions does the wheel make during that 60.0 s?

•24 A vinyl record is played by rotating the record so that an ap-

proximately circular groove in the vinyl slides under a stylus.

Bumps in the groove run into the stylus, causing it to oscillate.The

equipment converts those oscillations to electrical signals and then

to sound. Suppose that a record turns at the rate of , the

groove being played is at a radius of 10.0 cm, and the bumps in the

groove are uniformly separated by 1.75 mm.At what rate (hits per

second) do the bumps hit the stylus?

••25 (a) What is the angular speed vabout the polar axis of

SSM

331

3 rev/min

288 CHAPTER 10 ROTATION

its angular acceleration in revolutions per minute-squared? (b) How

many revolutions does the engine make during this 12 s interval?

••13 A flywheel turns through 40 rev as it slows from an

angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angu-

lar acceleration, find the time for it to come to rest. (b) What is its

angular acceleration? (c) How much time is required for it to com-

plete the first 20 of the 40 revolutions?

••14 A disk rotates about its central axis starting from rest and

accelerates with constant angular acceleration.At one time it is ro-

tating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s.

Calculate (a) the angular acceleration, (b) the time required to

complete the 60 revolutions, (c) the time required to reach the

10 rev/s angular speed, and (d) the number of revolutions from rest

until the time the disk reaches the 10 rev/s angular speed.

••15 Starting from rest, a wheel has constant a= 3.0 rad/s2.

During a certain 4.0 s interval, it turns through 120 rad. How much

time did it take to reach that 4.0 s interval?

••16 A merry-go-round rotates from rest with an angular accel-

eration of 1.50 rad/s2. How long does it take to rotate through

(a) the first 2.00 rev and (b) the next 2.00 rev?

••17 At t0, a flywheel has an angular velocity of 4.7 rad/s, a

constant angular acceleration of 0.25 rad/s2, and a reference line

at u00. (a) Through what maximum angle umax will the reference

line turn in the positive direction? What are the (b) first and

(d) negative time and (e) positive time will the reference line be

at 10.5 rad? (f) Graph uversus t, and indicate your answers.

•••18 A pulsar is a rapidly rotating neutron star that emits a radio

beam the way a lighthouse emits a light beam. We receive a radio

pulse for each rotation of the star.The period Tof rotation is found

by measuring the time between pulses.The pulsar in the Crab neb-

ula has a period of rotation of T0.033 s that is increasing at the

rate of 1.26 105s/y. (a) What is the pulsar’s angular acceleration

a? (b) If ais constant, how many years from now will the pulsar

stop rotating? (c) The pulsar originated in a supernova explosion

seen in the year 1054.Assuming constant a, find the initial T.

Module 10-3 Relating the Linear and Angular Variables

•19 What are the magnitudes of (a) the angular velocity, (b) the ra-

dial acceleration, and (c) the tangential acceleration of a spaceship

taking a circular turn of radius 3220 km at a speed of 29 000 km/h?

•20 An object rotates about a fixed axis, and the angular posi-

tion of a reference line on the object is given by u0.40e2t, where

uis in radians and tis in seconds. Consider a point on the object

that is 4.0 cm from the axis of rotation.At t0, what are the mag-

nitudes of the point’s (a) tangential component of acceleration

and (b) radial component of acceleration?

•21 Between 1911 and 1990, the top of the leaning bell

tower at Pisa, Italy, moved toward the south at an average rate of

1.2 mm/y. The tower is 55 m tall. In radians per second, what is the

average angular speed of the tower’s top about its base?

•22 An astronaut is tested in a centrifuge with radius 10 m and

rotating according to u0.30t2. At t5.0 s, what are the magni-

tudes of the (a) angular velocity, (b) linear velocity, (c) tangential

acceleration, and (d) radial acceleration?

•23 A flywheel with a diameter of 1.20 m is rotating

at an angular speed of 200 rev/min. (a) What is the angular speed

of the flywheel in radians per second? (b) What is the linear speed

of a point on the rim of the flywheel? (c) What constant angular ac-

WWWSSM

u

u1

2umax

SSM

ILW

Figure 10-31 Problem 28.

Light

beam

Light

source

Rotating

slotted wheel

Mirror

perpendicular

to light beam

Figure 10-32 Problem 29.

289

PROBLEMS

••41 In Fig. 10-37, two particles,

each with mass m0.85 kg, are fas-

tened to each other, and to a rotation

axis at O, by two thin rods, each with

length d5.6 cm and mass M

1.2 kg. The combination rotates

around the rotation axis with the an-

gular speed v0.30 rad/s. Measured

about O, what are the combination’s

(a) rotational inertia and (b) kinetic energy?

••42 The masses and coordinates of four particles are as

follows: 50 g, x2.0 cm, y2.0 cm; 25 g, x0, y4.0 cm; 25 g,

x3.0 cm, y3.0 cm; 30 g, x2.0 cm, y4.0 cm. What

are the rotational inertias of this collection about the (a) x, (b) y,

and (c) zaxes? (d) Suppose that we symbolize the answers to (a)

and (b) as Aand B, respectively. Then what is the answer to (c)

in terms of Aand B?



light passes through one of the slots at the outside edge of the

wheel, travels to a distant mirror, and returns to the wheel just in

time to pass through the next slot in the wheel. One such slotted

wheel has a radius of 5.0 cm and 500 slots around its edge.

Measurements taken when the mirror is L500 m from the

wheel indicate a speed of light of 3.0 105km/s. (a) What is the

(constant) angular speed of the wheel? (b) What is the linear

speed of a point on the edge of the wheel?

••30 A gyroscope flywheel of radius 2.83 cm is accelerated from

rest at 14.2 rad/s2until its angular speed is 2760 rev/min. (a) What is

the tangential acceleration of a point on the rim of the flywheel during

this spin-up process? (b) What is the radial acceleration of this point

when the flywheel is spinning at full speed? (c) Through what distance

does a point on the rim move during the spin-up?

••31 A disk, with a radius of 0.25 m, is to be rotated like a merry-

go-round through 800 rad, starting from rest, gaining angular speed

at the constant rate a1through the first 400 rad and then losing an-

gular speed at the constant rate a1until it is again at rest.The mag-

nitude of the centripetal acceleration of any portion of the disk is

not to exceed 400 m/s2. (a) What is the least time required for the ro-

tation? (b) What is the corresponding value of a1?

••32 A car starts from rest and moves around a circular track of

radius 30.0 m. Its speed increases at the constant rate of 0.500 m/s2.

(a) What is the magnitude of its net linear acceleration 15.0 s later?

(b) What angle does this net acceleration vector make with the

car’s velocity at this time?

Module 10-4 Kinetic Energy of Rotation

•33 Calculate the rota-

tional inertia of a wheel that has

a kinetic energy of 24 400 J when

rotating at 602 rev/min.

•34 Figure 10-33 gives angu-

lar speed versus time for a thin

rod that rotates around one

end. The scale on the vaxis is

set by (a) What

is the magnitude of the rod’s an-

gular acceleration? (b) At t

4.0 s, the rod has a rotational ki-

netic energy of 1.60 J.What is its kinetic energy at t0?

Module 10-5 Calculating the Rotational Inertia

•35 Two uniform solid cylinders, each rotating about its cen-

tral (longitudinal) axis at 235 rad/s, have the same mass of 1.25 kg but

differ in radius.What is the rotational kinetic energy of (a) the smaller

cylinder, of radius 0.25 m,and (b) the larger cylinder,of radius 0.75 m?

•36 Figure 10-34ashows a disk that can rotate about an axis at

SSM



vs6.0 rad/s.

SSM

a radial distance hfrom the center of the disk. Figure 10-34bgives

the rotational inertia Iof the disk about the axis as a function of that

distance h, from the center out to the edge of the disk. The scale on

the Iaxis is set by and What is

the mass of the disk?

•37 Calculate the rotational inertia of a meter stick, with

mass 0.56 kg, about an axis perpendicular to the stick and located

at the 20 cm mark. (Treat the stick as a thin rod.)

•38 Figure 10-35 shows three

0.0100 kg particles that have been

glued to a rod of length L6.00 cm

and negligible mass. The assembly

can rotate around a perpendicular

axis through point Oat the left end.

If we remove one particle (that is,

33% of the mass), by what percent-

age does the rotational inertia of the assembly around the rotation

axis decrease when that removed particle is (a) the innermost one

and (b) the outermost one?

••39 Trucks can be run on energy stored in a rotating flywheel,

with an electric motor getting the flywheel up to its top speed of

200prad/s. Suppose that one such flywheel is a solid, uniform

cylinder with a mass of 500 kg and a radius of 1.0 m. (a) What is the

kinetic energy of the flywheel after charging? (b) If the truck uses

an average power of 8.0 kW, for how many minutes can it operate

between chargings?

••40 Figure 10-36 shows an arrangement of 15 identical disks that

have been glued together in a rod-like shape of length L1.0000 m

and (total) mass M100.0 mg.The disks are uniform, and the disk

arrangement can rotate about a perpendicular axis through its cen-

tral disk at point O. (a) What is the rotational inertia of the

arrangement about that axis? (b) If we approximated the arrange-

ment as being a uniform rod of mass Mand length L, what percent-

age error would we make in using the formula in Table 10-2eto cal-

culate the rotational inertia?

SSM

IB0.150 kgm2.IA0.050 kgm2

0123456

t (s)

◊(rad/s)

Figure 10-33 Problem 34.

I (kg • m2)

IA0 0.1

h(m)

0.2

(b)(a)

Axis

Figure 10-34 Problem 36.

Figure 10-36 Problem 40.

Figure 10-35 Problems

38 and 62.

Axis

d d d

m m

Rotation axis

Figure 10-37 Problem 41.

290 CHAPTER 10 ROTATION

••53 Figure 10-43 shows a uniform

disk that can rotate around its center like a

merry-go-round. The disk has a radius of

2.00 cm and a mass of 20.0 grams and is ini-

tially at rest. Starting at time t0, two

forces are to be applied tangentially to the

rim as indicated, so that at time t1.25 s

the disk has an angular velocity of 250

rad/s counterclockwise. Force

has a magnitude of 0.100 N. What

is magnitude F2?

••54 In a judo foot-sweep

move, you sweep your opponent’s

left foot out from under him while

pulling on his gi (uniform) toward

that side. As a result, your oppo-

nent rotates around his right foot

and onto the mat. Figure 10-44

shows a simplified diagram of

your opponent as you face him,

with his left foot swept out. The

rotational axis is through point O.

The gravitational force on him

effectively acts at his center of

mass, which is a horizontal dis-

tance d28 cm from point O.His

mass is 70 kg, and his rotational in-

ertia about point Ois 65 kgm2.What is the magnitude of his initial

angular acceleration about point Oif your pull on his gi is (a) neg-

ligible and (b) horizontal with a magnitude of 300 N and applied at

height h1.4 m?

••55 In Fig. 10-45a, an irregularly shaped plastic plate with

uniform thickness and density (mass per unit volume) is to be

rotated around an axle that is perpendicular to the plate face

and through point O. The rotational inertia of the plate about

••43 The uniform solid

block in Fig. 10-38 has mass 0.172 kg

and edge lengths a3.5 cm, b8.4

cm, and c1.4 cm. Calculate its rota-

tional inertia about an axis through

one corner and perpendicular to the

large faces.

••44 Four identical particles of

mass 0.50 kg each are placed at the

vertices of a 2.0 m 2.0 m square

and held there by four massless rods, which form the sides of the

square. What is the rotational inertia of this rigid body about an

axis that (a) passes through the midpoints of opposite sides and

lies in the plane of the square, (b) passes through the midpoint of

one of the sides and is perpendicular to the plane of the square,

and (c) lies in the plane of the square and passes through two di-

agonally opposite particles?

WWWSSM rest, block 2 falls 75.0 cm in 5.00 s without the cord slipping on

the pulley. (a) What is the magnitude of the acceleration of the

blocks? What are (b) tension and (c) tension ? (d) What is

the magnitude of the pulley’s angular acceleration? (e) What is

its rotational inertia?

••52 In Fig. 10-42, a cylinder having a mass of 2.0 kg can rotate

about its central axis through point O. Forces are applied as shown:

F16.0 N, F24.0 N, F32.0 N, and F45.0 N. Also, r5.0 cm

and R12 cm. Find the (a) magnitude and (b) direction of the an-

gular acceleration of the cylinder. (During the rotation, the forces

maintain their same angles relative to the cylinder.)

Rotat

axis

Figure 10-38 Problem 43.

Figure 10-43

Problem 53.

Figure 10-42 Problem 52.

F4R

Rotation

axis

Figure 10-39 Problem 45.

Module 10-6 Torque

•45 The body in

Fig. 10-39 is pivoted at O, and

two forces act on it as shown. If

r11.30 m, r22.15 m, F1

4.20 N, F24.90 N, u175.0,

and u260.0, what is the net

torque about the pivot?

•46 The body in Fig. 10-40 is

pivoted at O. Three forces act

on it: FA10 N at point A, 8.0

m from O;FB16 N at B, 4.0

m from O; and FC19 N at C,

3.0 m from O. What is the net

torque about O?

•47 A small ball of mass

0.75 kg is attached to one end

of a 1.25-m-long massless rod,

SSM

ILWSSM

and the other end of the rod is hung from a pivot.When the resulting

pendulum is 30from the vertical, what is the magnitude of the gravi-

tational torque calculated about the pivot?

•48 The length of a bicycle pedal arm is 0.152 m, and a down-

ward force of 111 N is applied to the pedal by the rider.What is the

magnitude of the torque about the pedal arm’s pivot when the arm

is at angle (a) 30, (b) 90, and (c) 180with the vertical?

Module 10-7 Newton’s Second Law for Rotation

•49 During the launch from a board, a diver’s angular

speed about her center of mass changes from zero to 6.20 rad/s in 220

ms. Her rotational inertia about her center of mass is 12.0 kg m2.

During the launch, what are the magnitudes of (a) her average angu-

lar acceleration and (b) the average external

torque on her from the board?

•50 If a 32.0 N m torque on a wheel

causes angular acceleration 25.0 rad/s2,

what is the wheel’s rotational inertia?

••51 In Fig. 10-41, block 1 has mass

, block 2 has mass ,

and the pulley, which is mounted on a hor-

izontal axle with negligible friction, has

radius . When released fromR5.00 cm

m2500 gm1460 g

ILWSSM

Figure 10-44 Problem 54.

com

FCC

160°

90° O

135°

Figure 10-40 Problem 46.

T1T2

Figure 10-41

Problems 51 and 83.

291

PROBLEMS

that axle is measured with the following

method. A circular disk of mass 0.500 kg and

radius 2.00 cm is glued to the plate, with its

center aligned with point O(Fig. 10-45b). A

string is wrapped around the edge of the disk

the way a string is wrapped around a top.

Then the string is pulled for 5.00 s. As a re-

sult, the disk and plate are rotated by a con-

stant force of 0.400 N that is applied by the

string tangentially to the edge of the disk.

The resulting angular speed is 114 rad/s.

What is the rotational inertia of the plate

about the axle?

••56 Figure 10-46 shows

particles 1 and 2, each of mass

m, fixed to the ends of a rigid

massless rod of length L1

L2, with L120 cm and L2

80 cm. The rod is held hori-

zontally on the fulcrum and then released. What are the magni-

tudes of the initial accelerations of (a) particle 1 and (b) particle 2?

•••57 A pulley, with a rotational inertia of 1.0 103kgm2about

its axle and a radius of 10 cm, is acted on by a force applied tangentially

at its rim.The force magnitude varies in time as F0.50t0.30t2, with

Fin newtons and tin seconds.The pulley is initially at rest.At t3.0 s

what are its (a) angular acceleration and (b) angular speed?

Module 10-8 Work and Rotational Kinetic Energy

•58 (a) If R12 cm, M400 g, and m50 g in Fig. 10-19, find

the speed of the block after it has descended 50 cm starting from

rest. Solve the problem using energy conservation principles.

(b) Repeat (a) with R5.0 cm.

•59 An automobile crankshaft transfers energy from the engine

to the axle at the rate of 100 hp (74.6 kW) when rotating at a

speed of 1800 rev/min. What torque (in newton-meters) does the

crankshaft deliver?

•60 A thin rod of length 0.75 m and mass 0.42 kg is suspended

freely from one end. It is pulled to one side and then allowed to swing

like a pendulum, passing through its lowest position with angular

speed 4.0 rad/s. Neglecting friction and air resistance, find (a) the

rod’s kinetic energy at its lowest position and (b) how far above that

position the center of mass rises.

•61 A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is

rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

(a) How much work must be done to stop it? (b) What is the

required average power?

••62 In Fig. 10-35, three 0.0100 kg particles have been glued to a

rod of length L6.00 cm and negligible mass and can rotate

around a perpendicular axis through point Oat one end. How

much work is required to change the rotational rate (a) from 0 to

20.0 rad/s, (b) from 20.0 rad/s to 40.0 rad/s, and (c) from 40.0 rad/s to

60.0 rad/s? (d) What is the slope of a plot of the assembly’s kinetic

energy (in joules) versus the square of its rotation rate (in radians-

squared per second-squared)?

••63 A meter stick is held vertically with one end on

the floor and is then allowed to fall. Find the speed of the other end

just before it hits the floor, assuming that the end on the floor does

not slip. (Hint: Consider the stick to be a thin rod and use the con-

servation of energy principle.)

ILWSSM

••64 A uniform cylinder of radius 10 cm and mass 20 kg is

mounted so as to rotate freely about a horizontal axis that is paral-

lel to and 5.0 cm from the central longitudinal axis of the cylinder.

(a) What is the rotational inertia of the cylinder about the axis of

rotation? (b) If the cylinder is released from rest with its central

longitudinal axis at the same height as the axis about which the

cylinder rotates, what is the angular speed of the cylinder as it

passes through its lowest position?

•••65 A tall, cylindrical chimney falls over when its base

is ruptured.Treat the chimney as a thin rod of length 55.0 m. At the

instant it makes an angle of 35.0with the vertical as it falls, what

are (a) the radial acceleration of the top, and (b) the tangential ac-

celeration of the top. (Hint: Use energy considerations, not a torque.)

•••66 A uniform spherical shell of mass M4.5 kg and radius

(b)

(a)

Axle

Plate

Disk

String

Figure 10-45

Problem 55.

1 2

L1L2

Figure 10-46 Problem 56.

Figure 10-47 Problem 66.

R8.5 cm can rotate about a vertical axis on frictionless bearings

(Fig. 10-47). A massless cord passes around the equator of the shell,

over a pulley of rotational inertia I3.0 103kgm2and radius

r5.0 cm, and is attached to a small object of mass m0.60 kg.

There is no friction on the pulley’s axle; the cord does not slip on

the pulley.What is the speed of the object when it has fallen 82 cm

after being released from rest? Use energy considerations.

M, R

I, r

•••67 Figure 10-48 shows a rigid as-

sembly of a thin hoop (of mass mand ra-

dius R0.150 m) and a thin radial rod

(of mass mand length L2.00R). The

assembly is upright, but if we give it a

slight nudge, it will rotate around a hori-

zontal axis in the plane of the rod and

hoop, through the lower end of the rod.

Assuming that the energy given to the

assembly in such a nudge is negligible,

what would be the assembly’s angular speed about the rotation axis

when it passes through the upside-down (inverted) orientation?

Additional Problems

68 Two uniform solid spheres have the same mass of 1.65 kg, but

one has a radius of 0.226 m and the other has a radius of 0.854 m.

Each can rotate about an axis through its center. (a) What is the

magnitude tof the torque required to bring the smaller sphere

from rest to an angular speed of 317 rad/s in 15.5 s? (b) What is the

magnitude Fof the force that must be

applied tangentially at the sphere’s

equator to give that torque? What are

the corresponding values of (c) tand

(d) Ffor the larger sphere?

69 In Fig. 10-49, a small disk of radius

r2.00 cm has been glued to the edge of

a larger disk of radius R4.00 cm so that

Figure 10-48 Problem 67.

Rotation

Hoop

Rod

Figure 10-49 Problem 69.

292 CHAPTER 10 ROTATION

the disks lie in the same plane. The disks can be rotated around a per-

pendicular axis through point Oat the center of the larger disk. The

disks both have a uniform density (mass per unit volume) of 1.40 

103kg/m3and a uniform thickness of 5.00 mm. What is the rota-

tional inertia of the two-disk assembly about the rotation axis

through O?

70 A wheel, starting from rest, rotates with a constant angular

acceleration of 2.00 rad/s2. During a certain 3.00 s interval, it turns

through 90.0 rad. (a) What is the angular velocity of the wheel at

the start of the 3.00 s interval? (b) How long has the wheel been

turning before the start of the 3.00 s interval?

71 In Fig. 10-50, two 6.20 kg

blocks are connected by a massless

string over a pulley of radius 2.40 cm

and rotational inertia 7.40 104

kgm2. The string does not slip on

the pulley; it is not known whether

there is friction between the table and

the sliding block; the pulley’s axis is

frictionless. When this system is re-

leased from rest, the pulley turns through 0.130 rad in 91.0 ms and the

acceleration of the blocks is constant. What are (a) the magnitude of

the pulley’s angular acceleration, (b) the magnitude of either block’s

acceleration,(c) string tension T1, and (d) string tension T2?

72 Attached to each end of a thin steel rod of length 1.20 m and

mass 6.40 kg is a small ball of mass 1.06 kg. The rod is constrained

to rotate in a horizontal plane about a vertical axis through its mid-

point.At a certain instant, it is rotating at 39.0 rev/s. Because of fric-

tion, it slows to a stop in 32.0 s.Assuming a constant retarding torque

due to friction, compute (a) the angular acceleration, (b) the retard-

ing torque, (c) the total energy transferred from mechanical energy

to thermal energy by friction, and (d) the number of revolutions ro-

tated during the 32.0 s. (e) Now suppose that the retarding torque is

known not to be constant. If any of the quantities (a), (b),(c), and (d)

can still be computed without additional information, give its value.

73 A uniform helicopter rotor blade is 7.80 m long, has a mass of

110 kg, and is attached to the rotor axle by a single bolt. (a) What is

the magnitude of the force on the bolt from the axle when the ro-

tor is turning at 320 rev/min? (Hint: For this calculation the blade

can be considered to be a point mass at its center of mass. Why?)

(b) Calculate the torque that must be applied to the rotor to bring

it to full speed from rest in 6.70 s. Ignore air resistance. (The blade

cannot be considered to be a point mass for this calculation. Why

not? Assume the mass distribution of a uniform thin rod.) (c) How

much work does the torque do on the blade in order for the blade

to reach a speed of 320 rev/min?

74 Racing disks. Figure 10-51 shows

two disks that can rotate about their

centers like a merry-go-round. At

time t0, the reference lines of the

two disks have the same orientation.

Disk Ais already rotating, with a con-

stant angular velocity of 9.5 rad/s.

Disk Bhas been stationary but now begins to rotate at a constant

angular acceleration of 2.2 rad/s2. (a) At what time twill the refer-

ence lines of the two disks momentarily have the same angular dis-

placement u? (b) Will that time tbe the first time since t0 that

the reference lines are momentarily aligned?

75 A high-wire walker always attempts to keep his center of

mass over the wire (or rope). He normally carries a long, heavy pole

SSM

to help: If he leans, say, to his right (his com moves to the right) and is

in danger of rotating around the wire, he moves the pole to his left

(its com moves to the left) to slow the rotation and allow himself

time to adjust his balance. Assume that the walker has a mass of

70.0 kg and a rotational inertia of about the wire.What is

the magnitude of his angular acceleration about the wire if his com is

5.0 cm to the right of the wire and (a) he carries no pole and (b) the

14.0 kg pole he carries has its com 10 cm to the left of the wire?

76 Starting from rest at t0, a wheel undergoes a constant an-

gular acceleration. When t2.0 s, the angular velocity of the

wheel is 5.0 rad/s.The acceleration continues until t20 s, when it

abruptly ceases. Through what angle does the wheel rotate in the

interval t0 to t40 s?

77 A record turntable rotating at slows down

and stops in 30 s after the motor is turned off. (a) Find its (con-

stant) angular acceleration in revolutions per minute-squared.

(b) How many revolutions does it make in this time?

78 A rigid body is made of three

identical thin rods, each with length

L0.600 m, fastened together in the

form of a letter H(Fig. 10-52). The

body is free to rotate about a hori-

zontal axis that runs along the length

of one of the legs of the H. The body

is allowed to fall from rest from a position in which the plane of the

His horizontal. What is the angular speed of the body when the

plane of the His vertical?

79 (a) Show that the rotational inertia of a solid cylinder of

mass Mand radius Rabout its central axis is equal to the rotational

inertia of a thin hoop of mass Mand radius about its central

axis. (b) Show that the rotational inertia Iof any given body of

mass Mabout any given axis is equal to the rotational inertia of an

equivalent hoop about that axis, if the hoop has the same mass M

and a radius kgiven by

The radius kof the equivalent hoop is called the radius of gyration

of the given body.

80 A disk rotates at constant angular acceleration, from angular

position u110.0 rad to angular position u270.0 rad in 6.00 s. Its

angular velocity at u2is 15.0 rad/s. (a) What was its angular velocity

at u1? (b) What is the angular acceleration? (c) At what angular

position was the disk initially at rest? (d) Graph uversus time tand

angular speed vversus tfor the disk, from the beginning of the

motion (let t0 then).

81 The thin uniform rod in Fig. 10-53 has

length 2.0 m and can pivot about a horizontal,

frictionless pin through one end. It is released

from rest at angle u40above the horizontal.

Use the principle of conservation of energy to

determine the angular speed of the rod as it

passes through the horizontal position.

82 George Washington Gale Ferris, Jr., a

civil engineering graduate from Rensselaer Polytechnic Institute,

built the original Ferris wheel for the 1893 World’s Columbian

Exposition in Chicago. The wheel, an astounding engineering con-

struction at the time, carried 36 wooden cars, each holding up to 60

passengers, around a circle 76 m in diameter.The cars were loaded 6

at a time, and once all 36 cars were full, the wheel made a complete

kAI

R/22

SSM

331

3 rev/min

SSM

15.0 kgm2

Figure 10-51 Problem 74.

kADi

Figure 10-52 Problem 78.

Figure 10-50 Problem 71.

Figure 10-53

Problem 81.

the rings are given in the following table. A tangential force of

magnitude 12.0 N is applied to the outer edge of the outer ring for

0.300 s.What is the change in the angular speed of the construction

during the time interval?

Ring Mass (kg) Inner Radius (m) Outer Radius (m)

1 0.120 0.0160 0.0450

2 0.240 0.0900 0.1400

87 In Fig. 10-55, a wheel of ra-

dius 0.20 m is mounted on a friction-

less horizontal axle. A massless cord

is wrapped around the wheel and at-

tached to a 2.0 kg box that slides on

a frictionless surface inclined at an-

gle u20with the horizontal. The

box accelerates down the surface at 2.0 m/s2. What is the rota-

tional inertia of the wheel about the axle?

88 A thin spherical shell has a radius of 1.90 m.An applied torque

of 960 Nm gives the shell an angular acceleration of 6.20 rad/s2

about an axis through the center of the shell. What are (a) the rota-

tional inertia of the shell about that axis and (b) the mass of the shell?

89 A bicyclist of mass 70 kg puts all his mass on each downward-

moving pedal as he pedals up a steep road. Take the diameter of

293

PROBLEMS

rotation at constant angular speed in about 2 min. Estimate the

amount of work that was required of the machinery to rotate the

passengers alone.

83 In Fig.10-41, two blocks,of mass m1400 g and m2600 g,are

connected by a massless cord that is wrapped around a uniform disk

of mass M500 g and radius R12.0 cm.The disk can rotate with-

out friction about a fixed horizontal axis through its center; the cord

cannot slip on the disk.The system is released from rest. Find (a) the

magnitude of the acceleration of the blocks, (b) the tension T1in the

cord at the left,and (c) the tension T2in the cord at the right.

84 At 714 A.M. on June 30, 1908, a huge explosion

the circle in which the pedals rotate to be 0.40 m, and determine

the magnitude of the maximum torque he exerts about the rota-

tion axis of the pedals.

90 The flywheel of an engine is rotating at 25.0 rad/s. When the

engine is turned off, the flywheel slows at a constant rate and stops

in 20.0 s. Calculate (a) the angular acceleration of the flywheel,

(b) the angle through which the flywheel rotates in stopping, and

91 In Fig. 10-19a, a wheel of radius 0.20 m is mounted on a fric-

tionless horizontal axis. The rotational inertia of the wheel about the

axis is 0.40 kg m2. A massless cord wrapped around the wheel’s cir-

cumference is attached to a 6.0 kg box. The system is released from

rest.When the box has a kinetic energy of 6.0 J, what are (a) the wheel’s

rotational kinetic energy and (b) the distance the box has fallen?

92 Our Sun is 2.3 104ly (light-years) from the center of our

Milky Way galaxy and is moving in a circle around that center at a

speed of 250 km/s. (a) How long does it take the Sun to make one

revolution about the galactic center? (b) How many revolutions has

the Sun completed since it was formed about 4.5 109years ago?

93 A wheel of radius 0.20 m

is mounted on a frictionless horizon-

tal axis. The rotational inertia of the

wheel about the axis is 0.050 kgm2.

A massless cord wrapped around

the wheel is attached to a 2.0 kg

block that slides on a horizontal frictionless surface. If a horizontal

force of magnitude P3.0 N is applied to the block as shown in

Fig. 10-56, what is the magnitude of the angular acceleration of the

wheel? Assume the cord does not slip on the wheel.

94 If an airplane propeller rotates at 2000 rev/min while the air-

plane flies at a speed of 480 km/h relative to the ground, what is the

linear speed of a point on the tip of the propeller, at radius 1.5 m, as

seen by (a) the pilot and (b) an observer on the ground? The plane’s

velocity is parallel to the propeller’s axis of rotation.

95 The rigid body shown in

Fig. 10-57 consists of three particles

connected by massless rods. It is to be

rotated about an axis perpendicular

to its plane through point P. If M

0.40 kg, a30 cm, and b50 cm,

how much work is required to take

the body from rest to an angular

speed of 5.0 rad/s?

96 Beverage engineering. The pull

tab was a major advance in the engi-

neering design of beverage contain-

ers.The tab pivots on a central bolt in the can’s top.When you pull

upward on one end of the tab, the other end presses downward on

a portion of the can’s top that has been scored. If you pull upward

with a 10 N force, what force magnitude acts on the scored section?

(You will need to examine a can with a pull tab.)

97 Figure 10-58 shows a propeller blade that rotates at

2000 rev/min about a perpendicular axis at point B. Point Ais at

the outer tip of the blade, at radial distance

1.50 m. (a) What is the difference in the

magnitudes aof the centripetal acceleration

of point Aand of a point at radial distance

0.150 m? (b) Find the slope of a plot of a

versus radial distance along the blade.

SSM

86 Figure 10-54 shows a flat construction of

two circular rings that have a common center and

are held together by three rods of negligible mass.

The construction, which is initially at rest, can

rotate around the common center (like a merry-

go-round), where another rod of negligible mass

lies. The mass, inner radius, and outer radius of

occurred above remote central Siberia, at latitude 61N and lon-

gitude 102E; the fireball thus created was the brightest flash

seen by anyone before nuclear weapons. The Tunguska Event,

which according to one chance witness “covered an enormous part

of the sky,” was probably the explosion of a stony asteroid about 140

m wide. (a) Considering only Earth’s rotation, determine how much

later the asteroid would have had to arrive to put the explosion

above Helsinki at longitude 25E. This would have obliterated the

city. (b) If the asteroid had, instead, been a metallic asteroid, it could

have reached Earth’s surface. How much later would such an

asteroid have had to arrive to put the impact in the Atlantic Ocean

at longitude 20W? (The resulting tsunamis would have wiped out

coastal civilization on both sides of the Atlantic.)

85 A golf ball is launched at an angle of 20to the horizontal,

with a speed of 60 m/s and a rotation rate of 90 rad/s.Neglecting air

drag, determine the number of revolutions the ball makes by the

time it reaches maximum height.

Figure 10-54

Problem 86.

Figure 10-56 Problem 93.

b b

2M2M

a a

Figure 10-57 Problem 95.

Figure 10-55 Problem 87.

Figure 10-58

Problem 97.

294 CHAPTER 10 ROTATION

98 A yo-yo-shaped device

mounted on a horizontal fric-

tionless axis is used to lift a 30 kg

box as shown in Fig. 10-59. The

outer radius Rof the device is

0.50 m, and the radius rof the

hub is 0.20 m. When a constant

horizontal force of magni-

tude 140 N is applied to a rope

wrapped around the outside of

the device, the box, which is sus-

pended from a rope wrapped

around the hub, has an upward

acceleration of magnitude 0.80

m/s2.What is the rotational iner-

tia of the device about its axis of rotation?

99 A small ball with mass 1.30 kg is mounted on one end of a rod

0.780 m long and of negligible mass. The system rotates in a horizon-

tal circle about the other end of the rod at 5010 rev/min. (a) Calculate

the rotational inertia of the system about the axis of rotation. (b)

There is an air drag of 2.30 102N on the ball, directed opposite its

motion. What torque must be applied to the system to keep it rotat-

ing at constant speed?

100 Two thin rods (each of mass 0.20

kg) are joined together to form a rigid

body as shown in Fig. 10-60. One of the

rods has length L10.40 m, and the

other has length L20.50 m. What is

the rotational inertia of this rigid body

about (a) an axis that is perpendicular

to the plane of the paper and passes

through the center of the shorter rod

and (b) an axis that is perpendicular to

the plane of the paper and passes

through the center of the longer rod?

101 In Fig. 10-61, four pul-

leys are connected by two

belts. Pulley A(radius 15 cm)

is the drive pulley, and it ro-

tates at 10 rad/s. Pulley B(ra-

dius 10 cm) is connected by

belt 1 to pulley A. Pulley B

(radius 5 cm) is concentric with

pulley Band is rigidly attached

to it. Pulley C(radius 25 cm) is

connected by belt 2 to pulley B.

Calculate (a) the linear speed of

a point on belt 1, (b) the angular

app

and three connecting rods, with and

. The balls may be treated as particles, and the connecting

rods have negligible mass. Determine the rotational kinetic energy

of the object if it has an angular speed of 1.2 rad/s about (a) an axis

that passes through point Pand is perpendicular to the plane of the

figure and (b) an axis that passes through point P, is perpendicular

to the rod of length 2L, and lies in the plane of the figure.

103 In Fig. 10-63, a thin uniform rod

(mass 3.0 kg, length 4.0 m) rotates

freely about a horizontal axis Athat is

perpendicular to the rod and passes

through a point at distance d1.0 m

from the end of the rod. The kinetic

energy of the rod as it passes through

the vertical position is 20 J. (a) What is

the rotational inertia of the rod about

axis A? (b) What is the (linear) speed

of the end Bof the rod as the rod

passes through the vertical position?

mentarily stop in its upward swing?

104 Four particles, each of mass,

0.20 kg, are placed at the vertices of a

square with sides of length 0.50 m.The

particles are connected by rods of neg-

ligible mass. This rigid body can rotate

in a vertical plane about a horizontal

axis Athat passes through one of the

particles. The body is released from

rest with rod AB horizontal (Fig. 10-64).

(a) What is the rotational inertia of the

body about axis A? (b) What is the an-

gular speed of the body about axis A

when rod AB swings through the verti-

cal position?

105 Cheetahs running at top speed have been reported at an as-

tounding 114 km/h (about 71 mi/h) by observers driving alongside

the animals. Imagine trying to measure a cheetah’s speed by keeping

your vehicle abreast of the animal while also glancing at your

speedometer, which is registering 114 km/h. You keep the vehicle a

constant 8.0 m from the cheetah, but the noise of the vehicle causes

the cheetah to continuously veer away from you along a circular

path of radius 92 m. Thus, you travel along a circular path of radius

100 m. (a) What is the angular speed of you and the cheetah around

the circular paths? (b) What is the linear speed of the cheetah along

its path? (If you did not account for the circular motion, you would

conclude erroneously that the cheetah’s speed is 114 km/h, and that

type of error was apparently made in the published reports.)

106 A point on the rim of a 0.75-m-diameter grinding wheel

changes speed at a constant rate from 12 m/s to 25 m/s in 6.2 s.

What is the average angular acceleration of the wheel?

107 A pulley wheel that is 8.0 cm in diameter has a 5.6-m-long

cord wrapped around its periphery. Starting from rest, the wheel is

given a constant angular acceleration of 1.5 rad/s2. (a) Through

what angle must the wheel turn for the cord to unwind com-

pletely? (b) How long will this take?

108 A vinyl record on a turntable rotates at 33 rev/min.

(a) What is its angular speed in radians per second? What is the

linear speed of a point on the record (b) 15 cm and (c) 7.4 cm from

the turntable axis?



30

M1.6 kg, L0.60 m,

Rope

Yo-yo-shaped

device

Rigid mount

Hub

Fapp

Figure 10-59 Problem 98.

BBelt 1

Drive

pulley

Belt 2

Figure 10-61 Problem 101.

Figure 10-60 Problem 100.

2L1

Figure 10-62

Problem 102.

Figure 10-63 Problem 103.

Figure 10-64 Problem 104.

A B

Rotation

axis

speed of pulley B, (c) the angular speed of pulley B, (d) the linear

speed of a point on belt 2, and (e) the angular speed of pulley C.(Hint:

If the belt between two pulleys does not slip, the linear speeds at the

rims of the two pulleys must be equal.)

102 The rigid object shown in Fig. 10-62 consists of three balls

CHAPTER 11

Rolling, Torque, and Angular Momentum

295295

11-1 ROLLING AS TRANSLATION AND ROTATION COMBINED

After reading this module, you should be able to . . .

11.01 Identify that smooth rolling can be considered as a

combination of pure translation and pure rotation.

11.02 Apply the relationship between the center-of-mass

speed and the angular speed of a body in smooth rolling.

●For a wheel of radius Rrolling smoothly,

vcom vR,

where vcom is the linear speed of the wheel’s center of mass

and vis the angular speed of the wheel about its center.

●The wheel may also be viewed as rotating instantaneously

about the point Pof the “road” that is in contact with the

wheel. The angular speed of the wheel about this point

is the same as the angular speed of the wheel about

its center.

Learning Objectives

Key Ideas

What Is Physics?

As we discussed in Chapter 10, physics includes the study of rotation.Arguably,

the most important application of that physics is in the rolling motion of wheels

and wheel-like objects. This applied physics has long been used. For example,

when the prehistoric people of Easter Island moved their gigantic stone statues

from the quarry and across the island, they dragged them over logs acting as

rollers. Much later, when settlers moved westward across America in the 1800s,

they rolled their possessions first by wagon and then later by train.Today, like it

or not, the world is filled with cars, trucks, motorcycles, bicycles, and other

rolling vehicles.

The physics and engineering of rolling have been around for so long that

you might think no fresh ideas remain to be developed. However, skateboards

and inline skates were invented and engineered fairly recently, to become huge

financial successes. Street luge is now catching on, and the self-righting Segway

(Fig. 11-1) may change the way people move around in large cities. Applying the

physics of rolling can still lead to surprises and rewards. Our starting point in

exploring that physics is to simplify rolling motion.

Rolling as Translation and Rotation Combined

Here we consider only objects that roll smoothly along a surface; that is, the

objects roll without slipping or bouncing on the surface. Figure 11-2 shows how

complicated smooth rolling motion can be: Although the center of the object

moves in a straight line parallel to the surface, a point on the rim certainly

does not. However, we can study this motion by treating it as a combination

of translation of the center of mass and rotation of the rest of the object around

that center.

Figure 11-1 The self-righting Segway Human

Transporter.

Justin Sullivan/Getty Images, Inc.

To see how we do this, pretend you are standing on a sidewalk watching the

bicycle wheel of Fig. 11-3 as it rolls along a street.As shown, you see the center of

mass Oof the wheel move forward at constant speed vcom. The point Pon the

street where the wheel makes contact with the street surface also moves forward

at speed vcom, so that Palways remains directly below O.

During a time interval t, you see both Oand Pmove forward by a distance s.

The bicycle rider sees the wheel rotate through an angle uabout the center of the

wheel, with the point of the wheel that was touching the street at the beginning

of tmoving through arc length s. Equation 10-17 relates the arc length sto the

rotation angle u:

suR, (11-1)

where Ris the radius of the wheel. The linear speed vcom of the center of the

wheel (the center of mass of this uniform wheel) is ds/dt. The angular speed vof

the wheel about its center is du/dt. Thus, differentiating Eq. 11-1 with respect to

time (with Rheld constant) gives us

vcom vR(smooth rolling motion). (11-2)

A Combination. Figure 11-4 shows that the rolling motion of a wheel is a

combination of purely translational and purely rotational motions. Figure 11-4a

shows the purely rotational motion (as if the rotation axis through the center

were stationary): Every point on the wheel rotates about the center with angular

speed v. (This is the type of motion we considered in Chapter 10.) Every point on

the outside edge of the wheel has linear speed vcom given by Eq. 11-2. Figure 11-4b

shows the purely translational motion (as if the wheel did not rotate at all): Every

point on the wheel moves to the right with speed vcom.

The combination of Figs. 11-4aand 11-4byields the actual rolling motion

of the wheel, Fig. 11-4c. Note that in this combination of motions, the portion

of the wheel at the bottom (at point P) is stationary and the portion at the top

296 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Figure 11-2 A time-exposure photo-

graph of a rolling disk. Small lights

have been attached to the disk, one

at its center and one at its edge.

The latter traces out a curve called

acycloid.

Richard Megna/Fundamental Photographs

vcom

v = 2vcom

v = –vcom +vcom = 0

vcom

v = vcom

v = –vcom

(c)Rolling motion (b)Pure translation (a)Pure rotation

ω ω

Figure 11-4 Rolling motion of a wheel as a combination of purely rotational motion and

purely translational motion. (a) The purely rotational motion:All points on the wheel

move with the same angular speed v. Points on the outside edge of the wheel all move

with the same linear speed vvcom.The linear velocities of two such points, at top (T)

and bottom (P) of the wheel, are shown. (b) The purely translational motion:All points on

the wheel move to the right with the same linear velocity .(c) The rolling motion of

the wheel is the combination of (a) and (b).

com

Figure 11-3 The center of mass Oof a rolling

wheel moves a distance sat velocity

while the wheel rotates through angle u.

The point Pat which the wheel makes

contact with the surface over which the

wheel rolls also moves a distance s.

com

P P

vcom

(at point T) is moving at speed 2vcom, faster than any other portion of the

wheel. These results are demonstrated in Fig. 11-5, which is a time exposure of

a rolling bicycle wheel.You can tell that the wheel is moving faster near its top

than near its bottom because the spokes are more blurred at the top than at

the bottom.

The motion of any round body rolling smoothly over a surface can be sepa-

rated into purely rotational and purely translational motions, as in Figs. 11-4a

and 11-4b.

Rolling as Pure Rotation

Figure 11-6 suggests another way to look at the rolling motion of a wheel—

namely, as pure rotation about an axis that always extends through the point

where the wheel contacts the street as the wheel moves. We consider the rolling

motion to be pure rotation about an axis passing through point Pin Fig.11-4cand

perpendicular to the plane of the figure. The vectors in Fig. 11-6 then represent

the instantaneous velocities of points on the rolling wheel.

Question: What angular speed about this new axis will a stationary observer as-

sign to a rolling bicycle wheel?

Answer: The same vthat the rider assigns to the wheel as she or he observes it

in pure rotation about an axis through its center of mass.

To verify this answer, let us use it to calculate the linear speed of the top of the

rolling wheel from the point of view of a stationary observer. If we call the

wheel’s radius R, the top is a distance 2Rfrom the axis through Pin Fig. 11-6, so

the linear speed at the top should be (using Eq. 11-2)

vtop (v)(2R)2(vR)2vcom,

in exact agreement with Fig. 11-4c. You can similarly verify the linear speeds

shown for the portions of the wheel at points Oand Pin Fig.11-4c.

297

11-1 ROLLING AS TRANSLATION AND ROTATION COMBINED

Figure 11-5 A photograph of a rolling

bicycle wheel.The spokes near the

wheel’s top are more blurred than

those near the bottom because the top

ones are moving faster,as Fig. 11-4c

shows.

Courtesy Alice Halliday

Checkpoint 1

The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When

the bicycle is moving, is the linear speed at the very top of the rear wheel greater than,

less than, or the same as that of the very top of the front wheel? (b) Is the angular speed

of the rear wheel greater than, less than, or the same as that of the front wheel?

Figure 11-6 Rolling can be viewed as pure

rotation, with angular speed v, about an

axis that always extends through P.The

vectors show the instantaneous linear

velocities of selected points on the rolling

wheel.You can obtain the vectors by

combining the translational and rotational

motions as in Fig.11-4.

Rotation axis at P

The Kinetic Energy of Rolling

Let us now calculate the kinetic energy of the rolling wheel as measured by the

stationary observer. If we view the rolling as pure rotation about an axis through

Pin Fig.11-6, then from Eq. 10-34 we have

(11-3)

in which vis the angular speed of the wheel and IPis the rotational inertia of the

wheel about the axis through P. From the parallel-axis theorem of Eq. 10-36

(IIcom Mh2), we have

IPIcom MR2, (11-4)

in which Mis the mass of the wheel, Icom is its rotational inertia about an axis

through its center of mass, and R(the wheel’s radius) is the perpendicular

distance h. Substituting Eq. 11-4 into Eq. 11-3, we obtain

and using the relation vcom vR(Eq. 11-2) yields

(11-5)

We can interpret the term as the kinetic energy associated with the

rotation of the wheel about an axis through its center of mass (Fig.11-4a), and the

term as the kinetic energy associated with the translational motion of the

wheel’s center of mass (Fig. 11-4b).Thus, we have the following rule:

2Mvcom

2Icomv2

K1

2Icomv21



com.

K1

2Icomv21



R2v2,

K1

2IPv2,

298 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

●A smoothly rolling wheel has kinetic energy

where Icom is the rotational inertia of the wheel about its cen-

ter of mass and Mis the mass of the wheel.

●If the wheel is being accelerated but is still rolling smoothly,

the acceleration of the center of mass is related to thea

com

K1

2Icomv21



com,

angular acceleration aabout the center with

acom aR.

●If the wheel rolls smoothly down a ramp of angle u, its

acceleration along an xaxis extending up the ramp is

acom, x g sin u

1Icom /MR2.

Key Ideas

A rolling object has two types of kinetic energy: a rotational kinetic energy

due to its rotation about its center of mass and a translational kinetic

2Icomv2)

energy due to translation of its center of mass.(1

2Mvcom

11-2 FORCES AND KINETIC ENERGY OF ROLLING

After reading this module, you should be able to . . .

11.03 Calculate the kinetic energy of a body in smooth rolling as

the sum of the translational kinetic energy of the center of mass

and the rotational kinetic energy around the center of mass.

11.04 Apply the relationship between the work done on a

smoothly rolling object and the change in its kinetic energy.

11.05 For smooth rolling (and thus no sliding), conserve me-

chanical energy to relate initial energy values to the values

at a later point.

11.06 Draw a free-body diagram of an accelerating body that is

smoothly rolling on a horizontal surface or up or down a ramp.

11.07 Apply the relationship between the center-of-mass

acceleration and the angular acceleration.

11.08 For smooth rolling of an object up or down a

ramp, apply the relationship between the object’s

acceleration, its rotational inertia, and the angle of

the ramp.

Learning Objectives

The Forces of Rolling

Friction and Rolling

If a wheel rolls at constant speed, as in Fig. 11-3, it has no tendency to slide at the

point of contact P, and thus no frictional force acts there. However, if a net force

acts on the rolling wheel to speed it up or to slow it, then that net force causes ac-

celeration of the center of mass along the direction of travel. It also causes

the wheel to rotate faster or slower, which means it causes an angular

acceleration a.These accelerations tend to make the wheel slide at P.Thus, a fric-

tional force must act on the wheel at Pto oppose that tendency.

If the wheel does not slide, the force is a static frictional force and the

motion is smooth rolling.We can then relate the magnitudes of the linear acceler-

ation and the angular acceleration aby differentiating Eq. 11-2 with respect

to time (with Rheld constant). On the left side, dvcom/dt is acom, and on the right

side dv/dt is a. So, for smooth rolling we have

acom aR(smooth rolling motion). (11-6)

If the wheel does slide when the net force acts on it, the frictional force that

acts at Pin Fig. 11-3 is a kinetic frictional force The motion then is not smooth

rolling, and Eq. 11-6 does not apply to the motion. In this chapter we discuss only

smooth rolling motion.

Figure 11-7 shows an example in which a wheel is being made to rotate faster

while rolling to the right along a flat surface, as on a bicycle at the start of a race.

The faster rotation tends to make the bottom of the wheel slide to the left at

point P. A frictional force at P, directed to the right, opposes this tendency to

slide. If the wheel does not slide, that frictional force is a static frictional force

(as shown), the motion is smooth rolling, and Eq. 11-6 applies to the motion.

(Without friction, bicycle races would be stationary and very boring.)

If the wheel in Fig. 11-7 were made to rotate slower, as on a slowing bicy-

cle, we would change the figure in two ways: The directions of the center-of-

mass acceleration and the frictional force at point Pwould now be to

the left.

Rolling Down a Ramp

Figure 11-8 shows a round uniform body of mass Mand radius Rrolling smoothly

down a ramp at angle u,along an xaxis.We want to find an expression for the body’s

com

299

11-2 FORCES AND KINETIC ENERGY OF ROLLING

Figure 11-7 A wheel rolls horizontally with-

out sliding while accelerating with linear

acceleration , as on a bicycle at the start

of a race.A static frictional force acts

on the wheel at P, opposing its tendency

to slide.

com

Pfs

acom

Figure 11-8 A round uniform body of radius Rrolls down a ramp.The forces that act on it

are the gravitational force F

g, a normal force F

N, and a frictional force f

spointing up the

ramp. (For clarity, vector F

Nhas been shifted in the direction it points until its tail is at the

center of the body.)

Fgcos

Fgsin

FNForces FN and Fg cos

merely balance.

Forces Fg sin and fs

determine the linear

acceleration down

the ramp.

The torque due to fs

determines the

angular acceleration

around the com.

acceleration acom,xdown the ramp.We do this by using Newton’s second law in both

its linear version (Fnet Ma) and its angular version (tnet Ia).

We start by drawing the forces on the body as shown in Fig.11-8:

1. The gravitational force on the body is directed downward. The tail of the

vector is placed at the center of mass of the body. The component along the

ramp is Fgsin u,which is equal to Mg sin u.

2. A normal force is perpendicular to the ramp. It acts at the point of

contact P, but in Fig. 11-8 the vector has been shifted along its direction until

its tail is at the body’s center of mass.

3. A static frictional force acts at the point of contact Pand is directed up

the ramp.(Do you see why? If the body were to slide at P, it would slide down the

ramp.Thus, the frictional force opposing the sliding must be up the ramp.)

We can write Newton’s second law for components along the xaxis in Fig. 11-8

(Fnet,xmax) as

fsMg sin uMacom,x. (11-7)

This equation contains two unknowns, fsand acom,x. (We should not assume that fs

is at its maximum value fs,max. All we know is that the value of fsis just right for

the body to roll smoothly down the ramp, without sliding.)

We now wish to apply Newton’s second law in angular form to the body’s ro-

tation about its center of mass. First, we shall use Eq. 10-41 to write the

torques on the body about that point. The frictional force has moment arm R

and thus produces a torque Rfs, which is positive because it tends to rotate the

body counterclockwise in Fig. 11-8. Forces and have zero moment arms

about the center of mass and thus produce zero torques. So we can write the an-

gular form of Newton’s second law (tnet Ia) about an axis through the body’s

center of mass as

RfsIcoma. (11-8)

This equation contains two unknowns, fsand a.

Because the body is rolling smoothly, we can use Eq. 11-6 (acom aR) to relate

the unknowns acom,xand a. But we must be cautious because here acom,xis negative

(in the negative direction of the xaxis) and ais positive (counterclockwise). Thus

we substitute acom,x/Rfor ain Eq.11-8.Then,solving for fs, we obtain

(11-9)

Substituting the right side of Eq. 11-9 for fsin Eq. 11-7, we then find

(11-10)

We can use this equation to find the linear acceleration acom,xof any body rolling

along an incline of angle uwith the horizontal.

Note that the pull by the gravitational force causes the body to come down

the ramp, but it is the frictional force that causes the body to rotate and thus roll.

If you eliminate the friction (by, say, making the ramp slick with ice or grease) or

arrange for Mg sin uto exceed fs,max, then you eliminate the smooth rolling and

the body slides down the ramp.

acom,x g sin u

1Icom/MR2.

fsIcom

acom,x

R2.

(trF)

300 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Checkpoint 2

Disks Aand Bare identical and roll across a floor with equal speeds.Then disk Arolls

up an incline, reaching a maximum height h, and disk Bmoves up an incline that is

identical except that it is frictionless.Is the maximum height reached by disk Bgreater

than, less than, or equal to h?

301

11-3 THE YO-YO

Doing so, substituting for Icom (from Table 10-2f), and

then solving for vcom give us

4.10 m/s. (Answer)

Note that the answer does not depend on Mor R.

(b) What are the magnitude and direction of the frictional

force on the ball as it rolls down the ramp?

KEY IDEA

Because the ball rolls smoothly, Eq. 11-9 gives the frictional

force on the ball.

Calculations: Before we can use Eq. 11-9, we need the

ball’s acceleration acom,xfrom Eq. 11-10:

Note that we needed neither mass Mnor radius Rto find

acom,x. Thus, any size ball with any uniform mass would have

this smoothly rolling acceleration down a 30.0ramp.

We can now solve Eq. 11-9 as

(Answer)

Note that we needed mass Mbut not radius R. Thus, the

frictional force on any 6.00 kg ball rolling smoothly down

a 30.0ramp would be 8.40 N regardless of the ball’s ra-

dius but would be larger for a larger mass.



5(6.00 kg)(3.50 m/s2)8.40 N.

fsIcom

acom,x

R2

5MR2acom,x

R2

5Macom,x

 (9.8 m/s2) sin 30.0

12

3.50 m/s2.

acom,x g sin u

1Icom/MR2 g sin u

12

5MR2/MR2

vcom 2(10

7)gh 2(10

7)(9.8 m/s2)(1.20 m)

5MR2

Sample Problem 11.01 Ball rolling down a ramp

A uniform ball, of mass M6.00 kg and radius R, rolls

smoothly from rest down a ramp at angle u30.0(Fig. 11-8).

(a) The ball descends a vertical height h1.20 m to reach the

bottom of the ramp.What is its speed at the bottom?

KEY IDEAS

The mechanical energy Eof the ball–Earth system is con-

served as the ball rolls down the ramp.The reason is that the

only force doing work on the ball is the gravitational force, a

conservative force. The normal force on the ball from the

ramp does zero work because it is perpendicular to the

ball’s path. The frictional force on the ball from the ramp

does not transfer any energy to thermal energy because the

ball does not slide (it rolls smoothly).

Thus, we conserve mechanical energy (EfEi):

KfUfKiUi, (11-11)

where subscripts fand irefer to the final values (at the

bottom) and initial values (at rest), respectively.The gravita-

tional potential energy is initially UiMgh (where Mis the

ball’s mass) and finally Uf0.The kinetic energy is initially

Ki0. For the final kinetic energy Kf, we need an addi-

tional idea: Because the ball rolls, the kinetic energy in-

volves both translation and rotation, so we include them

both by using the right side of Eq. 11-5.

Calculations: Substituting into Eq. 11-11 gives us

(11-12)

where Icom is the ball’s rotational inertia about an axis

through its center of mass, vcom is the requested speed at the

bottom, and vis the angular speed there.

Because the ball rolls smoothly, we can use Eq. 11-2 to

substitute vcom/Rfor vto reduce the unknowns in Eq. 11-12.

2Icomv21



vcom

2)00Mgh,



Additional examples, video, and practice available at WileyPLUS

11-3 THE YO-YO

After reading this module, you should be able to . . .

11.09 Draw a free-body diagram of a yo-yo moving up or

down its string.

11.10 Identify that a yo-yo is effectively an object that rolls

smoothly up or down a ramp with an incline angle of 90.

11.11 For a yo-yo moving up or down its string, apply the rela-

tionship between the yo-yo’s acceleration and its rotational

inertia.

11.12 Determine the tension in a yo-yo’s string as the yo-yo

moves up or down its string.

●A yo-yo, which travels vertically up or down a string, can be treated as a wheel rolling along an inclined plane at angle u90.

Learning Objectives

Key Idea

The Yo-Yo

A yo-yo is a physics lab that you can fit in your pocket. If a yo-yo rolls down its

string for a distance h, it loses potential energy in amount mgh but gains kinetic

energy in both translational and rotational forms. As it climbs

back up, it loses kinetic energy and regains potential energy.

In a modern yo-yo, the string is not tied to the axle but is looped around it.

When the yo-yo “hits” the bottom of its string, an upward force on the axle from

the string stops the descent. The yo-yo then spins, axle inside loop, with only

rotational kinetic energy. The yo-yo keeps spinning (“sleeping”) until you “wake

it” by jerking on the string, causing the string to catch on the axle and the yo-yo to

climb back up. The rotational kinetic energy of the yo-yo at the bottom of its

string (and thus the sleeping time) can be considerably increased by throwing the

yo-yo downward so that it starts down the string with initial speeds vcom and vin-

stead of rolling down from rest.

To find an expression for the linear acceleration acom of a yo-yo rolling down

a string, we could use Newton’s second law (in linear and angular forms) just as

we did for the body rolling down a ramp in Fig. 11-8.The analysis is the same ex-

cept for the following:

1. Instead of rolling down a ramp at angle uwith the horizontal, the yo-yo rolls

down a string at angle u90with the horizontal.

2. Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of

radius R0(Fig. 11-9a).

3. Instead of being slowed by frictional force , the yo-yo is slowed by the force

on it from the string (Fig.11-9b).

The analysis would again lead us to Eq. 11-10. Therefore, let us just change the

notation in Eq. 11-10 and set u90to write the linear acceleration as

(11-13)

where Icom is the yo-yo’s rotational inertia about its center and Mis its mass. A

yo-yo has the same downward acceleration when it is climbing back up.

acom  g

1Icom/MR2

2Icomv2)(1

2Mvcom

302 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Figure 11-9 (a) A yo-yo, shown in cross

section. The string, of assumed negligible

thickness, is wound around an axle of

radius R0.(b) A free-body diagram for the

falling yo-yo. Only the axle is shown.

(a) (b)

11-4 TORQUE REVISITED

After reading this module, you should be able to . . .

11.13 Identify that torque is a vector quantity.

11.14 Identify that the point about which a torque is

calculated must always be specified.

11.15 Calculate the torque due to a force on a particle by

taking the cross product of the particle’s position vector

and the force vector, in either unit-vector notation or

magnitude-angle notation.

11.16 Use the right-hand rule for cross products to find the

direction of a torque vector.

●In three dimensions, torque is a vector quantity defined

relative to a fixed point (usually an origin); it is

where is a force applied to a particle and is a

position vector locating the particle relative to the fixed

point.

:r

:F

:●The magnitude of is given by

where fis the angle between and , is the component

of perpendicular to , and is the moment arm of .

●The direction of is given by the right-hand rule for cross

products.

r



trF sin frF

rF,

Learning Objectives

Key Ideas

Torque Revisited

In Chapter 10 we defined torque tfor a rigid body that can rotate around a fixed

axis. We now expand the definition of torque to apply it to an individual particle

that moves along any path relative to a fixed point (rather than a fixed axis). The

path need no longer be a circle, and we must write the torque as a vector that may

have any direction. We can calculate the magnitude of the torque with a formula

and determine its direction with the right-hand rule for cross products.

Figure 11-10ashows such a particle at point Ain an xy plane.A single force

in that plane acts on the particle, and the particle’s position relative to the origin

Ois given by position vector .The torque acting on the particle relative to the

fixed point Ois a vector quantity defined as

(torque defined). (11-14)

We can evaluate the vector (or cross) product in this definition of by using

the rules in Module 3-3. To find the direction of , we slide the vector (without

changing its direction) until its tail is at the origin O, so that the two vectors in the

vector product are tail to tail as in Fig. 11-10b. We then use the right-hand rule in

Fig. 3-19a, sweeping the fingers of the right hand from (the first vector in the

product) into (the second vector). The outstretched right thumb then gives the

direction of . In Fig.11-10b, it is in the positive direction of the zaxis.t

:r

:F

303

11-4 TORQUE REVISITED

Checkpoint 3

The position vector of a particle points along the positive direction of a zaxis. If

the torque on the particle is (a) zero,(b) in the negative direction of x,and (c) in the

negative direction of y,in what direction is the force causing the torque?

To determine the magnitude of , we apply the general result of Eq. 3-27

(cab sin f), finding trF sin f, (11-15)

where fis the smaller angle between the directions of and when the vectors

are tail to tail. From Fig.11-10b, we see that Eq. 11-15 can be rewritten as

(11-16)

where is the component of perpendicular to r

:. From Fig. 11-10c,

we see that Eq. 11-15 can also be rewritten as

(11-17)

where is the moment arm of (the perpendicular distance

between Oand the line of action of F

:).

r(r sin f)

trF,

(F sin f)

trF

,

Figure 11-10 Defining torque.(a) A force , lying in an xy plane, acts on a particle at point A.(b) This force produces a torque

on the particle with respect to the origin O.By the right-hand rule for vector (cross) products,the torque vector

points in the positive direction of z.Its magnitude is given by in (b) and by in (c).rFrF



(r

:F

(= r× F)

F (redrawn, with

tail at origin)

(b)

Line of action of F

(c)

(a)

Cross r into F.

Torque is in the

positive z direction.

304 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

the position vector are to see.) In Fig. 11-11d, the angle be-

tween the directions of and is 90and the symbol F

Sample Problem 11.02 Torque on a particle due to a force

In Fig. 11-11a, three forces, each of magnitude 2.0 N, act on a

particle. The particle is in the xz plane at point Agiven by

position vector , where r3.0 m and u30. What is the

torque, about the origin O, due to each force?

KEY IDEA

Because the three force vectors do not lie in a plane, we

must use cross products, with magnitudes given by Eq. 11-15

(trF sin f) and directions given by the right-hand rule.

Calculations: Because we want the torques with respect to

the origin O, the vector required for each cross product

is the given position vector. To determine the angle fbe-

tween and each force, we shift the force vectors of Fig.11-

11a, each in turn, so that their tails are at the origin. Figures

11-11b,c, and d, which are direct views of the xz plane, show

the shifted force vectors and , respectively. (Note

how much easier the angles between the force vectors and

2,F

Figure 11-11 (a) A particle at point Ais acted on by three forces,each parallel to a coordinate axis.The angle f(used in finding torque) is shown

(b) for and (c) for .(d) Torque t

3is perpendicular to both and (force is directed into the plane of the figure).(e) The torques.F

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(a)

F3F1

= 30°

(b)

1= 150°

OF1

Cross into F

. Torque

is into the figure (negative y).

= 30°

2= 120°

(c)F2

Cross into

Torque 2 is out of

the figure (positive y).

F2.

= 30°

(d)

(e)

The three

torques.

Cross into

Torque 3 is

in the xz plane.

F3.

means is directed into the page. (For out of the page, weF

would use .)

Now, applying Eq. 11-15, we find

t1rF1sin f1(3.0 m)(2.0 N)(sin 150)3.0 Nm,

t2rF2sin f2(3.0 m)(2.0 N)(sin 120)5.2 Nm,

and t3rF3sin f3(3.0 m)(2.0 N)(sin 90)

6.0 Nm. (Answer)

Next, we use the right-hand rule, placing the fingers of

the right hand so as to rotate into through the smaller of

the two angles between their directions. The thumb points in

the direction of the torque.Thus t

1is directed into the page in

Fig. 11-11b;t

2is directed out of the page in Fig. 11-11c; and

3is directed as shown in Fig. 11-11d. All three torque vec-

tors are shown in Fig.11-11e.

305

11-5 ANGULAR MOMENTUM

After reading this module, you should be able to . . .

11.17 Identify that angular momentum is a vector quantity.

11.18 Identify that the fixed point about which an angular

momentum is calculated must always be specified.

11.19 Calculate the angular momentum of a particle by taking

the cross product of the particle’s position vector and its

momentum vector, in either unit-vector notation or

magnitude-angle notation.

11.20 Use the right-hand rule for cross products to find the

direction of an angular momentum vector.

●The angular momentum of a particle with linear momen-

tum , mass m, and linear velocity is a vector quantity

defined relative to a fixed point (usually an origin) as

●The magnitude of is given by

rprmv,

rprmv

rmv sin f



:r

:p

:m(r

:v



:where fis the angle between and , and are

the components of and perpendicular to , and is

the perpendicular distance between the fixed point and

the extension of .

●The direction of is given by the right-hand rule: Position

your right hand so that the fingers are in the direction of .

Then rotate them around the palm to be in the direction of .

Your outstretched thumb gives the direction of .



r

v

p

Learning Objectives

Key Ideas

Angular Momentum

Recall that the concept of linear momentum and the principle of conservation

of linear momentum are extremely powerful tools. They allow us to predict

the outcome of, say, a collision of two cars without knowing the details of the col-

lision. Here we begin a discussion of the angular counterpart of , winding up in

Module 11-8 with the angular counterpart of the conservation principle, which

can lead to beautiful (almost magical) feats in ballet, fancy diving,ice skating, and

many other activities.

Figure 11-12 shows a particle of mass mwith linear momentum as

it passes through point Ain an xy plane. The angular momentum of this parti-

cle with respect to the origin Ois a vector quantity defined as

(angular momentum defined), (11-18)

where is the position vector of the particle with respect to O. As the particle

moves relative to Oin the direction of its momentum , position vector

rotates around O. Note carefully that to have angular momentum about O, the

particle does not itself have to rotate around O. Comparison of Eqs. 11-14 and 11-18

shows that angular momentum bears the same relation to linear momentum that

torque does to force. The SI unit of angular momentum is the kilogram-

meter-squared per second (kg m2/s), equivalent to the joule-second (Js).

Direction. To find the direction of the angular momentum vector in Fig. 11-

12, we slide the vector until its tail is at the origin O.Then we use the right-hand

rule for vector products, sweeping the fingers from into . The outstretched

thumb then shows that the direction of is in the positive direction of the zaxis in

Fig.11-12.This positive direction is consistent with the counterclockwise rotation of

position vector about the zaxis, as the particle moves. (A negative direction of

would be consistent with a clockwise rotation of about the zaxis.)

Magnitude. To find the magnitude of ,we use the general result of Eq. 3-27

to write (11-19)

where fis the smaller angle between and when these two vectors are tailp

rmv sin f,



(mv

)



:r

:p

:m(r

:v

)



:(mv

Figure 11-12 Defining angular momentum.A

particle passing through point Ahas linear

momentum , with the vector

lying in an xy plane.The particle has angular

momentum with respect to the

origin O. By the right-hand rule,the angular

momentum vector points in the positive

direction of z.(a) The magnitude of is

given by .(b) The magni-rprmv



:(r

:p

:(mv

(= r

p (redrawn, with

tail at origin)

Extension of p

(a)

(b)



tude of is also given by .rprmv

to tail. From Fig.11-12a, we see that Eq. 11-19 can be rewritten as

(11-20)

where is the component of perpendicular to and is the component

of perpendicular to . From Fig. 11-12b, we see that Eq. 11-19 can also be

rewritten as

(11-21)

where is the perpendicular distance between Oand the extension of .

Important. Note two features here: (1) angular momentum has meaning

only with respect to a specified origin and (2) its direction is always perpendicu-

lar to the plane formed by the position and linear momentum vectors and .p

r

rprmv,

v

p

rprmv,

306 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Checkpoint 4

In part aof the figure, particles 1 and 2 move around point Oin circles

with radii 2 m and 4 m. In part b, particles 3 and 4 travel along straight

lines at perpendicular distances of 4 m and 2 m from point O. Particle 5

moves directly away from O.All five particles have the same mass and

the same constant speed. (a) Rank the particles according to the magni-

tudes of their angular momentum about point O, greatest first. (b)

Which particles have negative angular momentum about point O?

(a)(b)

around Oas particle 1 moves. Thus, the angular momen-

tum vector for particle 1 is

Similarly, the magnitude of is

and the vector product is into the page, which is the

negative direction, consistent with the clockwise rotation of

around Oas particle 2 moves. Thus, the angular momen-

tum vector for particle 2 is

The net angular momentum for the two-particle system is

(Answer)

The plus sign means that the system’s net angular momen-

tum about point Ois out of the page.

2.0 kgm2/s.

L1210 kgm2/s (8.0 kgm2/s)

28.0 kgm2/s.

:p2

8.0 kgm2/s,

2r2p2(4.0 m)(2.0 kgm/s)



110 kgm2/s.

Sample Problem 11.03 Angular momentum of a two-particle system

Figure 11-13 shows an overhead view of two particles moving

at constant momentum along horizontal paths. Particle 1, with

momentum magnitude p15.0 kgm/s, has position vector

and will pass 2.0 m from point O. Particle 2, with momentum

magnitude p22.0 kgm/s, has position vector and will pass

4.0 m from point O. What are the magnitude and direction of

the net angular momentum about point Oof the two-

particle system?

KEY IDEA

To find , we can first find the individual angular momenta

and and then add them.To evaluate their magnitudes,

we can use any one of Eqs. 11-18 through 11-21. However,

Eq. 11-21 is easiest, because we are given the perpendicular

distances and and the momen-

tum magnitudes p1and p2.

Calculations: For particle 1, Eq. 11-21 yields

To find the direction of vector , we use Eq. 11-18 and the

right-hand rule for vector products. For , the vector

product is out of the page, perpendicular to the plane of

Fig. 11-13. This is the positive direction, consistent with the

counterclockwise rotation of the particle’s position vector

:p1



10 kgm2/s.

1r1p1(2.0 m)(5.0 kgm/s)

r2 (4.0 m)r1 (2.0 m)



Figure 11-13 Two particles pass

near point O.

r⊥1

r⊥2

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307

11-6 NEWTON’S SECOND LAW IN ANGULAR FORM

After reading this module, you should be able to . . .

11.21 Apply Newton’s second law in angular form to relate the torque acting on a particle to the resulting rate of change of the

particle’s angular momentum, all relative to a specified point.

●Newton’s second law for a particle can be written in angular form as

where is the net torque acting on the particle and is the angular momentum of the particle.

net

net d

dt ,

Learning Objective

Key Idea

Newton’s Second Law in Angular Form

Newton’s second law written in the form

(single particle) (11-22)

expresses the close relation between force and linear momentum for a single

particle. We have seen enough of the parallelism between linear and angular

quantities to be pretty sure that there is also a close relation between torque

and angular momentum. Guided by Eq. 11-22, we can even guess that it must be

(single particle). (11-23)

Equation 11-23 is indeed an angular form of Newton’s second law for a single particle:

net d

net dp

The (vector) sum of all the torques acting on a particle is equal to the time rate of

change of the angular momentum of that particle.

Equation 11-23 has no meaning unless the torques and the angular momentum

are defined with respect to the same point, usually the origin of the coordinate

system being used.

Proof of Equation 11-23

We start with Eq. 11-18, the definition of the angular momentum of a particle:

where is the position vector of the particle and is the velocity of the particle.

Differentiating* each side with respect to time tyields

(11-24)

However, is the acceleration of the particle, and is its velocity .

Thus, we can rewrite Eq. 11-24 as

d

dt m(r

:a

:v

:v

/dta

/dt

d

dt m



:dv

dt dr

dt v





:m(r

:v

:),



*In differentiating a vector product, be sure not to change the order of the two quantities (here and

) that form that product. (See Eq. 3-25.)

308 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Checkpoint 5

The figure shows the position vector of a particle

at a certain instant, and four choices for the direc-

tion of a force that is to accelerate the particle.All

four choices lie in the xy plane. (a) Rank the

choices according to the magnitude of the time rate

of change they produce in the angular mo-

mentum of the particle about point O, greatest

first. (b) Which choice results in a negative rate of change about O?

(d

:/dt)

on a particle and the angular momentum of the particle

are calculated around the same point, then the torque is re-

lated to angular momentum by Eq. 11-23 ( ).

Calculations: In order to use Eq. 11-18 to find the angular

momentum about the origin, we first must find an expres-

sion for the particle’s velocity by taking a time derivative of

its position vector. Following Eq. 4-10 ( ), we write

with in meters per second.

Next, let’s take the cross product of and using the

template for cross products displayed in Eq. 3-27:

Here the generic is and the generic is . However,

because we really don’t want to do more work than

needed, let’s first just think about our substitutions into

:b

:(aybzbyaz)i

ˆ(azbxbzax)j

ˆ(axbybxay)k

ˆ .

(4.00t1.00)i

:d

dt ((2.00t2t)i

ˆ5.00j

ˆ)

dr

/dtv

d

/dtt

Sample Problem 11.04 Torque and the time derivative of angular momentum

Figure 11-14ashows a freeze-frame of a 0.500 kg particle

moving along a straight line with a position vector given by

with in meters and tin seconds, starting at t0. The posi-

tion vector points from the origin to the particle. In unit-vector

notation, find expressions for the angular momentum of

the particle and the torque acting on the particle, both

with respect to (or about) the origin. Justify their algebraic

signs in terms of the particle’s motion.

KEY IDEAS

(1) The point about which an angular momentum of a par-

ticle is to be calculated must always be specified. Here it is

the origin. (2) The angular momentum of a particle is

given by Eq. 11-18 . (3) The sign

associated with a particle’s angular momentum is set by the

sense of rotation of the particle’s position vector (around

the rotation axis) as the particle moves: clockwise is nega-

tive and counterclockwise is positive. (4) If the torque acting

(

:r

:p

:m(r

:v

:))



:(2.00t2t)i

ˆ5.00j

Now (the vector product of any vector with itself is zero because the

angle between the two vectors is necessarily zero). Thus, the last term of this

expression is eliminated and we then have

We now use Newton’s second law to replace with its equal, the

vector sum of the forces that act on the particle, obtaining

(11-25)

Here the symbol indicates that we must sum the vector products for all

the forces. However, from Eq. 11-14, we know that each one of those vector prod-

ucts is the torque associated with one of the forces.Therefore, Eq. 11-25 tells us that

This is Eq. 11-23, the relation that we set out to prove.

net d

dt .

:F



d

dt r

:F

net 



:F

:).

net ma

d

dt m(r

:a

:)r

:ma

:v

:0

309

11-6 NEWTON’S SECOND LAW IN ANGULAR FORM

Figure 11-14 (a) A particle moving in a straight line, shown at time

t0. (b) The position vector at t0,1.00 s, and 2.00 s. (c) The first

step in applying the right-hand rule for cross products.(d) The sec-

ond step. (e) The angular momentum vector and the torque vector

are along the zaxis,which extends out of the plane of the figure.

y(m)

(a)

x(m)

y(m)

t 2 s t 1 s t 0

10 3

(b)

x(m)

r1r0

(c)

(d)

(e)



Both angular momentum

and torque point out of

figure, in the positive z

direction.

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the generic cross product. Because lacks any zcomponent

and because lacks any yor zcomponent, the only nonzero

term in the generic cross product is the very last one .

So,let’s cut to the (mathematical) chase by writing

Note that, as always, the cross product produces a vector

that is perpendicular to the original vectors.

To finish up Eq. 11-18, we multiply by the mass, finding

(Answer)

The torque about the origin then immediately follows from

Eq. 11-23:

(Answer)

which is in the positive direction of the zaxis.

Our result for tells us that the angular momentum is

in the positive direction of the zaxis. To make sense of that

positive result in terms of the rotation of the position vector,



10.0k

ˆ kgm2/s210.0k

ˆ Nm,

:d

dt (10.0t2.50)k

ˆ kgm2/s

(10.0t2.50)k

ˆ kgm2/s.



:(0.500 kg)[(20.0t5.00)k

ˆ m2/s]

:v

:(4.00t1.00)(5.00)k

ˆ(20.0t5.00)k

ˆ m2/s.

(bxay)k

let’s evaluate that vector for several times:

t0,

t1.00 s,

t2.00 s,

By drawing these results as in Fig. 11-14b, we see that ro-

tates counterclockwise in order to keep up with the particle.

That is the positive direction of rotation. Thus, even though

the particle is moving in a straight line, it is still moving

counterclockwise around the origin and thus has a positive

angular momentum.

We can also make sense of the direction of by applying

the right-hand rule for cross products (here , or, if you

like, , which gives the same direction). For any mo-

ment during the particle’s motion, the fingers of the right

hand are first extended in the direction of the first vector in

the cross product ( ) as indicated in Fig. 11-14c. The orienta-

tion of the hand (on the page or viewing screen) is then ad-

justed so that the fingers can be comfortably rotated about

the palm to be in the direction of the second vector in the

cross product ( ) as indicated in Fig. 11-14d.The outstretched

thumb then points in the direction of the result of the cross

product. As indicated in Fig. 11-14e, the vector is in the posi-

tive direction of the zaxis (which is directly out of the plane

of the figure), consistent with our previous result. Figure

11-14ealso indicates the direction of , which is also in the

positive direction of the zaxis because the angular momen-

tum is in that direction and is increasing in magnitude.

:v



10.0i

ˆ5.00j

ˆ mr

2

3.00i

ˆ5.00j

ˆ mr

1

5.00j

ˆ mr

0

The Angular Momentum of a System of Particles

Now we turn our attention to the angular momentum of a system of particles with

respect to an origin. The total angular momentum of the system is the (vector)

sum of the angular momenta of the individual particles (here with label i):

(11-26)

With time, the angular momenta of individual particles may change because

of interactions between the particles or with the outside.We can find the resulting

change in by taking the time derivative of Eq. 11-26.Thus,

(11-27)

From Eq. 11-23, we see that is equal to the net torque on the ith

particle.We can rewrite Eq. 11-27 as

(11-28)

That is, the rate of change of the system’s angular momentum is equal to the

vector sum of the torques on its individual particles. Those torques include inter-

nal torques (due to forces between the particles) and external torques (due to

forces on the particles from bodies external to the system). However, the forces

between the particles always come in third-law force pairs so their torques sum to

zero. Thus, the only torques that can change the total angular momentum of

the system are the external torques acting on the system.

Net External Torque. Let represent the net external torque, the vector

sum of all external torques on all particles in the system. Then we can write

Eq. 11-28 as

(system of particles), (11-29)



net dL



net

dt 



i1



net,i.



net,i

d

i/dt

dt 



i1

di

dt .

:

1

2

3   

n



i1



310 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

11-7 ANGULAR MOMENTUM OF A RIGID BODY

After reading this module, you should be able to . . .

11.22 For a system of particles, apply Newton’s second law

in angular form to relate the net torque acting on the

system to the rate of the resulting change in the system’s

angular momentum.

11.23 Apply the relationship between the angular momentum

of a rigid body rotating around a fixed axis and the body’s

rotational inertia and angular speed around that axis.

11.24 If two rigid bodies rotate about the same axis, calculate

their total angular momentum.

●The angular momentum of a system of particles is

the vector sum of the angular momenta of the individual

particles:

●The time rate of change of this angular momentum is equal

to the net external torque on the system (the vector sum of

:

1

2

n



i1



:the torques due to interactions of the particles of the system

with particles external to the system):

(system of particles).

●For a rigid body rotating about a fixed axis, the component

of its angular momentum parallel to the rotation axis is

LIv(rigid body, fixed axis).

net dL

Learning Objectives

Key Ideas

311

11-7 ANGULAR MOMENTUM OF A RIGID BODY

The net external torque acting on a system of particles is equal to the time

rate of change of the system’s total angular momentum .L



net

Equation 11-29 is analogous to (Eq. 9-27) but requires extra

caution:Torques and the system’s angular momentum must be measured relative

to the same origin. If the center of mass of the system is not accelerating relative

to an inertial frame, that origin can be any point. However, if it is accelerating,

then it must be the origin. For example, consider a wheel as the system of parti-

cles. If it is rotating about an axis that is fixed relative to the ground, then the ori-

gin for applying Eq. 11-29 can be any point that is stationary relative to the

ground. However, if it is rotating about an axis that is accelerating (such as when

it rolls down a ramp), then the origin can be only at its center of mass.

The Angular Momentum of a Rigid Body

Rotating About a Fixed Axis

We next evaluate the angular momentum of a system of particles that form a rigid

body that rotates about a fixed axis. Figure 11-15ashows such a body.The fixed axis

of rotation is a zaxis, and the body rotates about it with constant angular speed v.

We wish to find the angular momentum of the body about that axis.

We can find the angular momentum by summing the zcomponents of the an-

gular momenta of the mass elements in the body. In Fig. 11-15a, a typical mass el-

ement, of mass mi, moves around the zaxis in a circular path.The position of the

mass element is located relative to the origin Oby position vector . The radius

of the mass element’s circular path is the perpendicular distance between the

element and the zaxis.

The magnitude of the angular momentum of this mass element, with

respect to O, is given by Eq. 11-19:

where piand viare the linear momentum and linear speed of the mass element,

and 90is the angle between and . The angular momentum vector for the

mass element in Fig. 11-15ais shown in Fig. 11-15b; its direction must be perpen-

dicular to those of and .

The z Components. We are interested in the component of that is parallel

to the rotation axis, here the zaxis.That zcomponent is

The zcomponent of the angular momentum for the rotating rigid body as a

whole is found by adding up the contributions of all the mass elements that make

up the body.Thus, because we may write

(11-30)

We can remove vfrom the summation here because it has the same value for all

points of the rotating rigid body.

The quantity in Eq. 11-30 is the rotational inertia Iof the body

about the fixed axis (see Eq. 10-33).Thus Eq.11-30 reduces to

LIv(rigid body, fixed axis). (11-31)



miri

v





i1

miri



Lz



i1

iz 



i1

mivir

i



i1

mi(vr

i)r

i

vvr,

iz i sin u(ri sin u)(mivi)rimivi.



i(ri)( pi)(sin 90)(ri)(mivi),



ri,

net dP

/dt

Figure 11-15 (a) A rigid body rotates about a

zaxis with angular speed v.A mass ele-

ment of mass miwithin the body moves

about the zaxis in a circle with radius .

The mass element has linear momentum

and it is located relative to the origin

Oby position vector .Here the mass

element is shown when is parallel to the

xaxis. (b) The angular momentum with

respect to O,of the mass element in (a).The

zcomponent is also shown.iz



ri

i

iz

(a)

(b)

which is Newton’s second law in angular form.It says:

Conservation of Angular Momentum

So far we have discussed two powerful conservation laws, the conservation of

energy and the conservation of linear momentum. Now we meet a third law of

this type, involving the conservation of angular momentum. We start from

We have dropped the subscript z, but you must remember that the angular

momentum defined by Eq. 11-31 is the angular momentum about the rotation

axis.Also, Iin that equation is the rotational inertia about that same axis.

Table 11-1, which supplements Table 10-3, extends our list of corresponding

linear and angular relations.

312 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Checkpoint 6

In the figure, a disk, a

hoop, and a solid sphere

are made to spin about

fixed central axes (like a

top) by means of strings

wrapped around them, with the strings producing the same constant tangential force

on all three objects.The three objects have the same mass and radius,and they are

initially stationary. Rank the objects according to (a) their angular momentum about

their central axes and (b) their angular speed, greatest first, when the strings have

been pulled for a certain time t.

Disk Hoop Sphere

F F F

11-8 CONSERVATION OF ANGULAR MOMENTUM

After reading this module, you should be able to . . .

11.25 When no external net torque acts on a system along a specified axis, apply the conservation of angular momentum to

relate the initial angular momentum value along that axis to the value at a later instant.

●The angular momentum of a system remains constant if the net external torque acting on the system is zero:

(isolated system)

or (isolated system).

This is the law of conservation of angular momentum.

iL

:a constant

Learning Objective

Key Idea

Table 11-1 More Corresponding Variables and Relations for Translational

and Rotational Motiona

Translational Rotational

Force Torque

Linear momentum Angular momentum

Linear momentumbAngular momentumbL

:(

i)P

:(p



: (r

:p

:)p



: (r

:F

:)F

Linear momentumbAngular momentumcLIv

Newton’s second lawbNewton’s second lawb

Conservation lawdConservation lawd

aSee also Table 10-3.

bFor systems of particles,including rigid bodies.

cFor a rigid body about a fixed axis,with Lbeing the component along that axis.

dFor a closed, isolated system.

:a constantP

:a constant



net dL

net dP

:Mv

com

313

11-8 CONSERVATION OF ANGULAR MOMENTUM

If the net external torque acting on a system is zero, the angular momentum of

the system remains constant, no matter what changes take place within the system.

Equations 11-32 and 11-33 are vector equations; as such, they are equivalent

to three component equations corresponding to the conservation of angular

momentum in three mutually perpendicular directions. Depending on the

torques acting on a system, the angular momentum of the system might be con-

served in only one or two directions but not in all directions:

Eq. 11-29 ,which is Newton’s second law in angular form. If no (t

net dL

:/dt)

net external torque acts on the system, this equation becomes or

(isolated system). (11-32)

This result, called the law of conservation of angular momentum, can also be

written as

or (isolated system). (11-33)

Equations 11-32 and 11-33 tell us:

iL



net angular momentum

at some initial time ti







net angular momentum

at some later time tf



:a constant

:/dt 0,

If the component of the net external torque on a system along a certain axis is

zero, then the component of the angular momentum of the system along that axis

cannot change, no matter what changes take place within the system.

This is a powerful statement: In this situation we are concerned with only the initial

and final states of the system; we do not need to consider any intermediate state.

We can apply this law to the isolated body in Fig. 11-15, which rotates around

the zaxis. Suppose that the initially rigid body somehow redistributes its mass

relative to that rotation axis, changing its rotational inertia about that axis.

Equations 11-32 and 11-33 state that the angular momentum of the body cannot

change. Substituting Eq. 11-31 (for the angular momentum along the rotational

axis) into Eq. 11-33, we write this conservation law as

IiviIfvf. (11-34)

Here the subscripts refer to the values of the rotational inertia Iand angular

speed vbefore and after the redistribution of mass.

Like the other two conservation laws that we have discussed, Eqs. 11-32 and

11-33 hold beyond the limitations of Newtonian mechanics. They hold for parti-

cles whose speeds approach that of light (where the theory of special relativity

reigns), and they remain true in the world of subatomic particles (where quantum

physics reigns). No exceptions to the law of conservation of angular momentum

have ever been found.

We now discuss four examples involving this law.

1. The spinning volunteer Figure 11-16 shows a student seated on a stool that

can rotate freely about a vertical axis. The student, who has been set into

rotation at a modest initial angular speed vi, holds two dumbbells in his

outstretched hands. His angular momentum vector lies along the vertical ro-

tation axis, pointing upward.

The instructor now asks the student to pull in his arms; this action reduces

his rotational inertia from its initial value Iito a smaller value Ifbecause he

moves mass closer to the rotation axis. His rate of rotation increases markedly,

Figure 11-16 (a) The student has a relatively

large rotational inertia about the rotation

axis and a relatively small angular speed.

(b) By decreasing his rotational inertia, the

student automatically increases his angular

speed.The angular momentum of the

rotating system remains unchanged.

Rotation axis

(a)

(b)

314 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

from vito vf.The student can then slow down by extending his arms once more,

moving the dumbbells outward.

No net external torque acts on the system consisting of the student, stool,

and dumbbells.Thus, the angular momentum of that system about the rotation

axis must remain constant, no matter how the student maneuvers the dumb-

bells. In Fig. 11-16a, the student’s angular speed viis relatively low and his ro-

tational inertia Iiis relatively high.According to Eq. 11-34, his angular speed in

Fig. 11-16bmust be greater to compensate for the decreased If.

2. The springboard diver Figure 11-17 shows a diver doing a forward one-and-a-

half-somersault dive. As you should expect, her center of mass follows a para-

bolic path. She leaves the springboard with a definite angular momentum

about an axis through her center of mass, represented by a vector pointing into

the plane of Fig. 11-17, perpendicular to the page. When she is in the air, no net

external torque acts on her about her center of mass, so her angular momentum

about her center of mass cannot change. By pulling her arms and legs into the

closed tuck position, she can considerably reduce her rotational inertia about the

same axis and thus, according to Eq. 11-34, considerably increase her angular

speed. Pulling out of the tuck position (into the open layout position) at the end

of the dive increases her rotational inertia and thus slows her rotation rate so she

can enter the water with little splash. Even in a more complicated dive involving

both twisting and somersaulting, the angular momentum of the diver must be

conserved,in both magnitude and direction, throughout the dive.

3. Long jump When an athlete takes off from the ground in a running long jump,

the forces on the launching foot give the athlete an angular momentum with a

forward rotation around a horizontal axis. Such rotation would not allow the

jumper to land properly: In the landing,the legs should be together and extended

forward at an angle so that the heels mark the sand at the greatest distance. Once

airborne, the angular momentum cannot change (it is conserved) because no ex-

ternal torque acts to change it. However,the jumper can shift most of the angular

momentum to the arms by rotating them in windmill fashion (Fig. 11-18). Then

the body remains upright and in the proper orientation for landing.

Figure 11-18 Windmill motion of the arms

during a long jump helps maintain body

orientation for a proper landing.

(a) (b)

Figure 11-19 (a) Initial phase of a tour jeté:

large rotational inertia and small angular

speed. (b) Later phase: smaller rotational

inertia and larger angular speed.

Figure 11-17 The diver’s angular momentum

is constant throughout the dive, being

represented by the tail of an arrow that

is perpendicular to the plane of the figure.

Note also that her center of mass (see the

dots) follows a parabolic path.

Her angular momentum

is fixed but she can still

control her spin rate.

4. Tour jeté In a tour jeté, a ballet performer leaps with a small twisting motion

on the floor with one foot while holding the other leg perpendicular to the

body (Fig. 11-19a).The angular speed is so small that it may not be perceptible

315

11-8 CONSERVATION OF ANGULAR MOMENTUM

Checkpoint 7

A rhinoceros beetle rides the rim of a small disk that rotates like a merry-go-round.

If the beetle crawls toward the center of the disk, do the following (each relative to

the central axis) increase, decrease, or remain the same for the beetle–disk system:

(a) rotational inertia, (b) angular momentum, and (c) angular speed?

to the audience. As the performer ascends, the outstretched leg is brought

down and the other leg is brought up, with both ending up at angle uto the

body (Fig. 11-19b). The motion is graceful, but it also serves to increase the

rotation because bringing in the initially outstretched leg decreases the per-

former’s rotational inertia. Since no external torque acts on the airborne

performer, the angular momentum cannot change. Thus, with a decrease in

rotational inertia, the angular speed must increase. When the jump is well

executed, the performer seems to suddenly begin to spin and rotates 180

before the initial leg orientations are reversed in preparation for the landing.

Once a leg is again outstretched, the rotation seems to vanish.

Sample Problem 11.05 Conservation of angular momentum, rotating wheel demo

Figure 11-20ashows a student, again sitting on a stool that

can rotate freely about a vertical axis. The student, initially

at rest, is holding a bicycle wheel whose rim is loaded with

lead and whose rotational inertia Iwh about its central axis is

1.2 kgm2. (The rim contains lead in order to make the value

of Iwh substantial.)

The wheel is rotating at an angular speed vwh of 3.9

rev/s; as seen from overhead, the rotation is counterclock-

wise. The axis of the wheel is vertical, and the angular

momentum of the wheel points vertically upward.

The student now inverts the wheel (Fig. 11-20b) so that,

as seen from overhead, it is rotating clockwise. Its angular

momentum is now . The inversion results in the stu-

dent, the stool, and the wheel’s center rotating together as a

composite rigid body about the stool’s rotation axis, with ro-

tational inertia Ib6.8 kgm2. (The fact that the wheel is

also rotating about its center does not affect the mass distri-

bution of this composite body; thus, Ibhas the same value

whether or not the wheel rotates.) With what angular speed

vband in what direction does the composite body rotate af-

ter the inversion of the wheel?

KEY IDEAS

1. The angular speed vbwe seek is related to the final angu-

lar momentum of the composite body about the

stool’s rotation axis by Eq.11-31 (LIv).

2. The initial angular speed vwh of the wheel is related to the

angular momentum of the wheel’s rotation about its

center by the same equation.

3. The vector addition of and gives the total angular

momentum of the system of the student, stool, and

wheel.

4. As the wheel is inverted, no net external torque acts on

tot



L

Figure 11-20 (a) A student holds a bicycle wheel rotating around a

vertical axis.(b) The student inverts the wheel, setting himself

into rotation. (c) The net angular momentum of the system must

remain the same in spite of the inversion.

(a) (b)

Lwh –Lwh

= +

(c)

Initial Final

The student now has

angular momentum,

and the net of these

two vectors equals

the initial vector.

that system to change about any vertical axis.

(Torques due to forces between the student and the

wheel as the student inverts the wheel are internal to the

system.) So, the system’s total angular momentum is con-

served about any vertical axis, including the rotation axis

through the stool.

tot

316 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Calculations: The conservation of is represented with

vectors in Fig. 11-20c. We can also write this conservation in

terms of components along a vertical axis as

Lb,f Lwh,f Lb,i Lwh,i, (11-35)

where iand frefer to the initial state (before inversion of

the wheel) and the final state (after inversion). Because

inversion of the wheel inverted the angular momentum

vector of the wheel’s rotation, we substitute Lwh,i for

Lwh,f. Then, if we set Lb,i 0 (because the student,

the stool, and the wheel’s center were initially at rest),

Eq. 11-35 yields

Lb,f 2Lwh,i.

tot

The rotational inertia of a disk rotating about its central

axis is given by Table 10-2cas . Substituting 6.00mfor

the mass M, our disk here has rotational inertia

. (11-36)

(We don’t have values for mand R, but we shall continue

with physics courage.)

From Eq. 10-33, we know that the rotational inertia of

the cockroach (a particle) is equal to mr2. Substituting the

cockroach’s initial radius ( ) and final radius

( ), we find that its initial rotational inertia about the

rotation axis is

(11-37)

and its final rotational inertia about the rotation axis is

. (11-38)

So, the cockroach–disk system initially has the rotational

inertia , (11-39)

and finally has the rotational inertia

. (11-40)

Next, we use Eq. 11-31 ( ) to write the fact that

the system’s final angular momentum Lfis equal to the sys-

tem’s initial angular momentum Li:

or .

After canceling the unknowns mand R, we come to

. (Answer)

Note that v decreased because part of the mass moved out-

ward, thus increasing that system’s rotational inertia.

vf1.37 rad/s

4.00mR2vf3.64mR2(1.50 rad/s)

IfvfIivi

LIv

IfIdIcf 4.00mR2

IiIdIci 3.64mR2

Icf mR2

Ici 0.64mR2

rR

r0.800R

Id3.00mR2

2MR2

Sample Problem 11.06 Conservation of angular momentum, cockroach on disk

In Fig.11-21, a cockroach with mass mrides on a disk of mass

6.00mand radius R. The disk rotates like a merry-go-round

around its central axis at angular speed The

cockroach is initially at radius , but then it crawls

out to the rim of the disk. Treat the cockroach as a particle.

What then is the angular speed?

KEY IDEAS

(1) The cockroach’s crawl changes the mass distribution (and

thus the rotational inertia) of the cockroach–disk system.

(2) The angular momentum of the system does not change

because there is no external torque to change it. (The forces

and torques due to the cockroach’s crawl are internal to the

system.) (3) The magnitude of the angular momentum of a

rigid body or a particle is given by Eq. 11-31 ( ).

Calculations: We want to find the final angular speed. Our

key is to equate the final angular momentum Lfto the initial

angular momentum Li, because both involve angular speed.

They also involve rotational inertia I.So, let’s start by finding

the rotational inertia of the system of cockroach and disk

before and after the crawl.

LIv

r0.800R

vi1.50 rad/s.

Additional examples, video, and practice available at WileyPLUS

Figure 11-21 A cockroach rides at radius ron a disk rotating like a

merry-go-round.

Rotation axis

Using Eq. 11-31, we next substitute Ibvbfor Lb,f and Iwhvwh

for Lwh,i and solve for vb, finding

(Answer)

This positive result tells us that the student rotates counter-

clockwise about the stool axis as seen from overhead. If the

student wishes to stop rotating, he has only to invert the

wheel once more.

(2)(1.2 kgm2)(3.9 rev/s)

6.8 kgm21.4 rev/s.



b2Iwh

vwh

317

11-9 PRECESSION OF A GYROSCOPE

Precession of a Gyroscope

A simple gyroscope consists of a wheel fixed to a shaft and free to spin about the

axis of the shaft. If one end of the shaft of a nonspinning gyroscope is placed on a

support as in Fig. 11-22aand the gyroscope is released, the gyroscope falls by rotat-

ing downward about the tip of the support. Since the fall involves rotation, it is gov-

erned by Newton’s second law in angular form,which is given by Eq. 11-29:

(11-41)

This equation tells us that the torque causing the downward rotation (the fall)

changes the angular momentum of the gyroscope from its initial value of zero.

The torque is due to the gravitational force acting at the gyroscope’s center

of mass, which we take to be at the center of the wheel.The moment arm relative to

the support tip,located at Oin Fig. 11-22a,is .The magnitude of is

tMgr sin 90Mgr (11-42)

(because the angle between and is 90), and its direction is as shown in

Fig. 11-22a.

A rapidly spinning gyroscope behaves differently. Assume it is released with

the shaft angled slightly upward. It first rotates slightly downward but then, while

it is still spinning about its shaft, it begins to rotate horizontally about a vertical

axis through support point Oin a motion called precession.

Why Not Just Fall Over? Why does the spinning gyroscope stay aloft instead

of falling over like the nonspinning gyroscope? The clue is that when the spinning

gyroscope is released, the torque due to must change not an initial angular mo-

mentum of zero but rather some already existing nonzero angular momentum due

to the spin.

To see how this nonzero initial angular momentum leads to precession, we first

consider the angular momentum of the gyroscope due to its spin. To simplify the

situation,we assume the spin rate is so rapid that the angular momentum due to pre-

cession is negligible relative to . We also assume the shaft is horizontal when pre-

cession begins, as in Fig.11-22b.The magnitude of is given by Eq. 11-31:

LIv, (11-43)

where Iis the rotational moment of the gyroscope about its shaft and vis the an-

gular speed at which the wheel spins about the shaft. The vector points along

the shaft, as in Fig. 11-22b. Since is parallel to torque must be

perpendicular to .L



:dL

dt .

Figure 11-22 (a) A nonspinning gyroscope

falls by rotating in an xz plane because of

torque . (b) A rapidly spinning gyroscope,

with angular momentum precesses

around the zaxis.Its precessional motion is

in the xy plane. (c) The change in

angular momentum leads to a rotation of

about O.

:/dt

=dL

___

(a)

(b)

L(t)

___

L(t+ dt)

Circular path

taken by head

of Lvector

(c)

Support

11-9 PRECESSION OF A GYROSCOPE

After reading this module, you should be able to . . .

11.26 Identify that the gravitational force acting on a spinning

gyroscope causes the spin angular momentum vector (and

thus the gyroscope) to rotate about the vertical axis in a

motion called precession.

11.27 Calculate the precession rate of a gyroscope.

11.28 Identify that a gyroscope’s precession rate is

independent of the gyroscope’s mass.

●A spinning gyroscope can precess about a vertical axis through its support at the rate

where Mis the gyroscope’s mass, ris the moment arm, Iis the rotational inertia, and vis the spin rate.

 Mgr

Iv,

Learning Objectives

Key Idea

318 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

M, mass cancels from Eq. 11-46; thus is independent of the mass.

Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle

to the horizontal. It holds as well for a spinning top, which is essentially a spinning

gyroscope at an angle to the horizontal.

According to Eq. 11-41, torque causes an incremental change in the

angular momentum of the gyroscope in an incremental time interval dt; that is,

(11-44)

However, for a rapidly spinning gyroscope, the magnitude of is fixed by

Eq. 11-43. Thus the torque can change only the direction of not its magnitude.

From Eq. 11-44 we see that the direction of is in the direction of per-

pendicular to . The only way that can be changed in the direction of

without the magnitude Lbeing changed is for to rotate around the zaxis as

shown in Fig. 11-22c. maintains its magnitude, the head of the vector follows

a circular path, and is always tangent to that path. Since must always

point along the shaft, the shaft must rotate about the zaxis in the direction of .

Thus we have precession. Because the spinning gyroscope must obey Newton’s

law in angular form in response to any change in its initial angular momentum, it

must precess instead of merely toppling over.

Precession. We can find the precession rate by first using Eqs. 11-44 and

11-42 to get the magnitude of :

dL tdt Mgr dt. (11-45)

As changes by an incremental amount in an incremental time interval dt, the shaft

and precess around the zaxis through incremental angle df. (In Fig. 11-22c, angle

dfis exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45, we find that dfis

given by

Dividing this expression by dt and setting the rate df/dt, we obtain

(precession rate). (11-46)

This result is valid under the assumption that the spin rate vis rapid. Note that 

decreases as vis increased. Note also that there would be no precession if the

gravitational force did not act on the gyroscope, but because Iis a function ofMg

 Mgr

dfdL

LMgr dt

Iv.

:,dL

:t

:dt.

Rolling Bodies For a wheel of radius Rrolling smoothly,

vcom vR, (11-2)

where vcom is the linear speed of the wheel’s center of mass and vis

the angular speed of the wheel about its center. The wheel may

also be viewed as rotating instantaneously about the point Pof the

“road” that is in contact with the wheel. The angular speed of the

wheel about this point is the same as the angular speed of

the wheel about its center.The rolling wheel has kinetic energy

(11-5)

where Icom is the rotational inertia of the wheel about its center of

mass and Mis the mass of the wheel. If the wheel is being accelerated

but is still rolling smoothly,the acceleration of the center of mass

is related to the angular acceleration aabout the center with

acom aR. (11-6)

com

K1

2Icomv21



com,

Review & Summary

If the wheel rolls smoothly down a ramp of angle u, its acceleration

along an xaxis extending up the ramp is

(11-10)

Torque as a Vector In three dimensions, torque is a vector

quantity defined relative to a fixed point (usually an origin); it is

(11-14)

where is a force applied to a particle and is a position vector lo-

cating the particle relative to the fixed point.The magnitude of is

(11-15, 11-16, 11-17)

where fis the angle between and is the component of

perpendicular to and is the moment arm of . The direction

of is given by the right-hand rule.t

r



trF sin frF

rF,

:r

:F

acom, x g sin u

1Icom /MR2.

319

QUESTIONS

Angular Momentum of a Particle The angular momentum

of a particle with linear momentum mass m, and linear velocity is

a vector quantity defined relative to a fixed point (usually an origin) as

(11-18)

The magnitude of is given by

(11-19)

(11-20)

(11-21)

where fis the angle between and and are the compo-

nents of and perpendicular to and is the perpendicular

distance between the fixed point and the extension of .The direc-

tion of is given by the right-hand rule for cross products.

Newton’s Second Law in Angular Form Newton’s second

law for a particle can be written in angular form as

(11-23)

where is the net torque acting on the particle and is the angu-

lar momentum of the particle.

Angular Momentum of a System of Particles The angu-

lar momentum of a system of particles is the vector sum of the

angular momenta of the individual particles:

(11-26)L

:

1

2   

n



i1



net

net d

dt ,



r

v

p

rprmv,

rprmv

rmv sin f



:r

:p

:m(r

:v



:The time rate of change of this angular momentum is equal to the

net external torque on the system (the vector sum of the torques

due to interactions with particles external to the system):

(system of particles). (11-29)

Angular Momentum of a Rigid Body For a rigid body

rotating about a fixed axis, the component of its angular

momentum parallel to the rotation axis is

LIv(rigid body, fixed axis). (11-31)

Conservation of Angular Momentum The angular mo-

mentum of a system remains constant if the net external torque

acting on the system is zero:

(isolated system) (11-32)

or (isolated system). (11-33)

This is the law of conservation of angular momentum.

Precession of a Gyroscope A spinning gyroscope can pre-

cess about a vertical axis through its support at the rate

(11-46)

where Mis the gyroscope’s mass, ris the moment arm, Iis the rota-

tional inertia, and vis the spin rate.

 Mgr

Iv,

iL

:a constant

net dL

1Figure 11-23 shows three particles

of the same mass and the same constant

speed moving as indicated by the veloc-

ity vectors. Points a,b,c, and dform a

square, with point eat the center. Rank

the points according to the magnitude

of the net angular momentum of the

three-particle system when measured

about the points,greatest first.

2Figure 11-24 shows two parti-

cles Aand Bat xyz coordinates

(1 m, 1 m, 0) and (1 m, 0, 1 m).

Acting on each particle are three

numbered forces, all of the same

magnitude and each directed paral-

lel to an axis. (a) Which of the

forces produce a torque about the

origin that is directed parallel to y?

(b) Rank the forces according to

the magnitudes of the torques they

produce on the particles about the ori-

gin, greatest first.

3What happens to the initially sta-

tionary yo-yo in Fig. 11-25 if you pull it

via its string with (a) force (the line

of action passes through the point of

contact on the table, as indicated),

(b) force (the line of action passesF

above the point of contact), and (c) force (the line of action

passes to the right of the point of contact)?

4The position vector of a particle relative to a certain point

has a magnitude of 3 m, and the force on the particle has a mag-

nitude of 4 N. What is the angle between the directions of and

if the magnitude of the associated torque equals (a) zero and (b) 12

Nm?

5In Fig. 11-26, three forces of the

same magnitude are applied to a par-

ticle at the origin ( acts directly into

the plane of the figure). Rank the

forces according to the magnitudes of

the torques they create about (a)

point P1,(b) point P2,and (c) point P3,

greatest first.

6The angular momenta of a

particle in four situations are (1)

; (2) ; (3) ; (4) . In which situation

is the net torque on the particle (a) zero, (b) positive and con-

stant, (c) negative and increasing in magnitude (t0), and (d)

negative and decreasing in magnitude (t0)?

7A rhinoceros beetle rides the rim of a horizontal disk rotating

counterclockwise like a merry-go-round. If the beetle then walks

along the rim in the direction of the rotation, will the magnitudes

of the following quantities (each measured about the rotation axis)

increase, decrease, or remain the same (the disk is still rotating in

the counterclockwise direction): (a) the angular momentum of the

4/t26t2

3t4

(t)

Figure 11-23 Question 1.

Figure 11-24 Question 2.

Figure 11-25 Question 3.

Questions

Figure 11-26 Question 5.

320 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

ing to the magnitude of the angular momentum of the particle

measured about them, greatest first.

beetle–disk system, (b) the angular momentum and angular veloc-

ity of the beetle, and (c) the angular momentum and angular velocity

of the disk? (d) What are your answers if the beetle walks in the di-

rection opposite the rotation?

8Figure 11-27 shows an overhead

view of a rectangular slab that can

spin like a merry-go-round about its

center at O. Also shown are seven

paths along which wads of bubble

gum can be thrown (all with the

same speed and mass) to stick onto

the stationary slab. (a) Rank the paths according to the angular

speed that the slab (and gum) will have after the gum sticks, great-

est first. (b) For which paths will the angular momentum of the slab

(and gum) about Obe negative from

the view of Fig.11-27?

9Figure 11-28 gives the angular mo-

mentum magnitude Lof a wheel ver-

sus time t. Rank the four lettered time

intervals according to the magnitude

of the torque acting on the wheel,

greatest first.

10 Figure 11-29 shows a particle moving at constant velocity

and five points with their xy coordinates. Rank the points accord-

11 A cannonball and a marble roll smoothly from rest down an

incline. Is the cannonball’s (a) time to the bottom and (b) transla-

tional kinetic energy at the bottom more than, less than, or the

same as the marble’s?

12 A solid brass cylinder and a solid wood cylinder have the

same radius and mass (the wood cylinder is longer). Released to-

gether from rest,they roll down an incline.(a) Which cylinder reaches

the bottom first, or do they tie? (b) The wood cylinder is then short-

ened to match the length of the brass cylinder, and the brass cylinder

is drilled out along its long (central) axis to match the mass of the

wood cylinder.Which cylinder now wins the race, or do they tie?

(–3, 1) (9, 1)

(1, 3) c

(–1, –2)

(4, –1)

A B C D

Figure 11-28 Question 9.

Figure 11-29 Question 10.

•5 A 1000 kg car has four 10 kg wheels. When the car is mov-

ing, what fraction of its total kinetic energy is due to rotation of the

wheels about their axles? Assume

that the wheels are uniform disks

of the same mass and size.Why do

you not need to know the radius

of the wheels?

••6 Figure 11-30 gives the speed

vversus time tfor a 0.500 kg ob-

ject of radius 6.00 cm that rolls

smoothly down a 30ramp. The

scale on the velocity axis is set by

vs 4.0 m/s. What is the rota-

tional inertia of the object?

••7 In Fig. 11-31, a solid cylin-

der of radius 10 cm and mass 12 kg

starts from rest and rolls without

slipping a distance L6.0 m down

a roof that is inclined at angle u

30. (a) What is the angular speed of

the cylinder about its center as it

leaves the roof? (b) The roof’s edge

is at height H5.0 m. How far hori-

zontally from the roof’s edge does

the cylinder hit the level ground?

ILW

Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign

SSM Worked-out solution available in Student Solutions Manual

•–••• Number of dots indicates level of problem difficulty

Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com

WWW Worked-out solution is at

ILW Interactive solution is at http://www.wiley.com/college/halliday

Problems

Module 11-1 Rolling as Translation and Rotation Combined

•1 A car travels at 80 km/h on a level road in the positive direction

of an xaxis. Each tire has a diameter of 66 cm. Relative to a woman

riding in the car and in unit-vector notation, what are the velocity

at the (a) center, (b) top, and (c) bottom of the tire and the magni-

tude aof the acceleration at the (d) center, (e) top, and (f) bottom

of each tire? Relative to a hitchhiker sitting next to the road and in

unit-vector notation, what are the velocity at the (g) center,

(h) top, and (i) bottom of the tire and the magnitude aof the

acceleration at the (j) center,(k) top, and (l) bottom of each tire?

•2 An automobile traveling at 80.0 km/h has tires of 75.0 cm di-

ameter.(a) What is the angular speed of the tires about their axles?

(b) If the car is brought to a stop uniformly in 30.0 complete turns

of the tires (without skidding), what is the magnitude of the angu-

lar acceleration of the wheels? (c) How far does the car move dur-

ing the braking?

Module 11-2 Forces and Kinetic Energy of Rolling

•3 A 140 kg hoop rolls along a horizontal floor so that the

hoop’s center of mass has a speed of 0.150 m/s. How much work

must be done on the hoop to stop it?

•4 A uniform solid sphere rolls down an incline. (a) What must be

the incline angle if the linear acceleration of the center of the

sphere is to have a magnitude of 0.10g? (b) If a frictionless block

were to slide down the incline at that angle, would its acceleration

magnitude be more than, less than, or equal to 0.10g? Why?

SSM

v (m/s)

00.2 0.4

t (s)

0.6 0.8 1

Figure 11-30 Problem 6.

Figure 11-31 Problem 7.

Figure 11-27 Question 8.

4 5 6

321

PROBLEMS

bowling ball of radius R11 cm

along a lane. The ball (Fig. 11-38)

slides on the lane with initial speed

vcom,0 8.5 m/s and initial angular

speed v00. The coefficient of ki-

netic friction between the ball and the lane is 0.21.The kinetic fric-

tional force acting on the ball causes a linear acceleration of the

ball while producing a torque that causes an angular acceleration

of the ball. When speed vcom has decreased enough and angular

speed vhas increased enough, the ball stops sliding and then rolls

smoothly. (a) What then is vcom in terms of v? During the sliding,

what are the ball’s (b) linear acceleration and (c) angular accelera-

tion? (d) How long does the ball slide? (e) How far does the ball

slide? (f) What is the linear speed of the ball when smooth rolling

begins?

•••16 Nonuniform cylindrical object. In Fig. 11-39, a cylindrical

object of mass Mand radius Rrolls smoothly from rest down a

ramp and onto a horizontal section. From there it rolls off the ramp

and onto the floor, landing a horizontal distance d0.506 m from

the end of the ramp. The initial height of the object is H0.90 m;

the end of the ramp is at height h0.10 m. The object consists of

an outer cylindrical shell (of a certain uniform density) that is

glued to a central cylinder (of a different uniform density).The ro-

tational inertia of the object can be expressed in the general form

IbMR2, but bis not 0.5 as it is for a cylinder of uniform density.

Determine b.

••8 Figure 11-32 shows the po-

tential energy U(x) of a solid

ball that can roll along an xaxis.

The scale on the Uaxis is set by

Us 100 J. The ball is uniform,

rolls smoothly, and has a mass of

0.400 kg. It is released at x7.0 m

headed in the negative direction

of the xaxis with a mechanical

energy of 75 J. (a) If the ball can

reach x0 m, what is its speed

there, and if it cannot, what is its

turning point? Suppose, instead, it is headed in the positive direc-

tion of the xaxis when it is released at x7.0 m with 75 J. (b) If

the ball can reach x13 m, what is

its speed there, and if it cannot, what

is its turning point?

••9 In Fig. 11-33, a solid ball

rolls smoothly from rest (starting at

height H6.0 m) until it leaves the

horizontal section at the end of the

track, at height h2.0 m. How far

horizontally from point Adoes the

ball hit the floor?

••10 A hollow sphere of radius 0.15 m, with rotational inertia

I0.040 kgm2about a line through its center of mass, rolls

without slipping up a surface inclined at 30to the horizontal. At

a certain initial position, the sphere’s total kinetic energy is 20 J.

(a) How much of this initial kinetic energy is rotational?

(b) What is the speed of the center of mass of the sphere at the

initial position? When the sphere has moved 1.0 m up the incline

from its initial position, what are (c) its total kinetic energy and

(d) the speed of its center of mass?

••11 In Fig. 11-34, a constant hori-

zontal force of magnitude 10 N is

applied to a wheel of mass 10 kg and

radius 0.30 m. The wheel rolls

smoothly on the horizontal surface,

and the acceleration of its center of

mass has magnitude 0.60 m/s2. (a) In

unit-vector notation, what is the fric-

tional force on the wheel? (b) What is the rotational inertia of the

wheel about the rotation axis through its center of mass?

app

rolls smoothly from rest down a ramp and onto a circular loop of

radius 0.48 m. The initial height of the ball is h0.36 m. At the

loop bottom, the magnitude of the normal force on the ball is

2.00Mg. The ball consists of an outer spherical shell (of a certain

uniform density) that is glued to a central sphere (of a different

uniform density). The rotational inertia of the ball can be ex-

pressed in the general form IbMR2, but bis not 0.4 as it is for a

ball of uniform density. Determine b.

•••14 In Fig. 11-37, a small, solid, uniform ball is to be shot

from point Pso that it rolls smoothly along a horizontal path, up

along a ramp, and onto a plateau.Then it leaves the plateau hori-

zontally to land on a game board, at a horizontal distance dfrom

the right edge of the plateau. The vertical heights are h15.00

cm and h21.60 cm. With what speed must the ball be shot at

point Pfor it to land at d6.00 cm?

Figure 11-32 Problem 8.

Figure 11-33 Problem 9.

02 4 6 8 101214

U (J)

x (m)

Figure 11-37 Problem 14.

Figure 11-36 Problem 13.

Figure 11-35 Problem 12.

hRQ

Ball

h1d

Figure 11-34 Problem 11.

Fapp

•••15 A bowler throws a

Figure 11-39 Problem 16.

fkvcom

Figure 11-38 Problem 15.

••12 In Fig. 11-35, a solid brass

ball of mass 0.280 g will roll

smoothly along a loop-the-loop

track when released from rest along

the straight section. The circular

loop has radius R14.0 cm, and the

ball has radius rR. (a) What is hif

the ball is on the verge of leaving

the track when it reaches the top of

the loop? If the ball is released at

height h6.00R, what are the (b)

magnitude and (c) direction of the

horizontal force component acting

on the ball at point Q?

•••13 Nonuniform ball. In Fig. 11-

36, a ball of mass Mand radius R

322 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

position vector .What

are (a) the torque on the particle about

the origin, in unit-vector notation, and (b)

the angle between the directions of

and ?

Module 11-5 Angular Momentum

•26 At the instant of Fig. 11-40, a 2.0 kg

particle Phas a position vector of magni-

tude 3.0 m and angle u145and a velocity

vector of magnitude 4.0 m/s and angle

u230. Force of magnitude 2.0 N andF

:(3.0 m)i

ˆ(4.0 m)j

(b) force with components F2x0,F2y2.0 N,F2z4.0 N?

••22 A particle moves through an xyz coordinate system while

a force acts on the particle. When the particle has the position

vector the force is given

by and the corresponding torque

about the origin is

Determine Fx.

••23 Force acts on a pebble with position

vector relative to the origin. In unit-vec-

tor notation, what is the resulting torque on the pebble about (a)

the origin and (b) the point (2.0 m, 0, 3.0 m)?

••24 In unit-vector notation, what is the torque about the origin

on a jar of jalapeño peppers located at coordinates (3.0 m, 2.0 m,

4.0 m) due to (a) force (b)

force and (c) the vector

sum of and ? (d) Repeat part (c) for the torque about the

point with coordinates (3.0 m, 2.0 m, 4.0 m).

••25 Force acts on a particle withF

:(8.0 N)i

ˆ(6.0 N)j

SSM

(5.0 N)k

ˆ,F

2(3.0 N)i

ˆ(4.0 N)j

ˆ

(4.0 N)j

ˆ(5.0 N)k

ˆ,F

1(3.0 N)i

ˆ

:(0.50 m)j

ˆ(2.0 m)k

:(2.0 N)i

ˆ(3.0 N)k

(2.00 Nm)j

ˆ(1.00 Nm)k

ˆ .

:(4.00 Nm)i

ˆ

:Fxi

ˆ(7.00 N)j

ˆ(6.00 N)k

:(2.00 m)i

ˆ(3.00 m)j

ˆ(2.00 m)k

ˆ,

Module 11-3 The Yo-Yo

•17 A yo-yo has a rotational inertia of 950 g cm2and

a mass of 120 g. Its axle radius is 3.2 mm, and its string is 120 cm

long. The yo-yo rolls from rest down to the end of the string.

(a) What is the magnitude of its linear acceleration? (b) How long

does it take to reach the end of the string? As it reaches the end of

the string, what are its (c) linear speed, (d) translational kinetic en-

ergy, (e) rotational kinetic energy, and (f) angular speed?

•18 In 1980, over San Francisco Bay, a large yo-yo was

released from a crane. The 116 kg yo-yo consisted of two uniform

disks of radius 32 cm connected by an axle of radius 3.2 cm. What

was the magnitude of the acceleration of the yo-yo during (a) its

fall and (b) its rise? (c) What was the tension in the cord on which

it rolled? (d) Was that tension near the cord’s limit of 52 kN?

Suppose you build a scaled-up version of the yo-yo (same shape

and materials but larger). (e) Will the magnitude of your yo-yo’s

acceleration as it falls be greater than, less than, or the same as that

of the San Francisco yo-yo? (f) How about the tension in the cord?

Module 11-4 Torque Revisited

•19 In unit-vector notation, what is the net torque about the ori-

gin on a flea located at coordinates (0, 4.0 m, 5.0 m) when forces

and act on the flea?

•20 A plum is located at coordinates (2.0 m, 0, 4.0 m). In unit-

vector notation, what is the torque about the origin on the plum if

that torque is due to a force whose only component is (a) Fx

6.0 N, (b) Fx6.0 N, (c) Fz6.0 N, and (d) Fz6.0 N?

•21 In unit-vector notation, what is the torque about the origin on

a particle located at coordinates (0, 4.0 m, 3.0 m) if that torque is

due to (a) force with components F1x2.0 N, F1yF1z0, andF

2(2.0 N)j

1(3.0 N)k

SSM

that has position vector and velocity vector

. About the origin and in unit-vector nota-

tion, what are (a) the object’s angular momentum and (b) the

torque acting on the object?

•28 A 2.0 kg particle-like object moves in a plane with velocity

components vx30 m/s and vy60 m/s as it passes through the

point with (x,y) coordinates of (3.0, 4.0) m. Just then, in unit-

vector notation, what is its angular momentum relative to (a) the

origin and (b) the point located at (2.0, 2.0) m?

•29 In the instant of Fig. 11-41,

two particles move in an xy plane.

Particle P1has mass 6.5 kg and

speed v12.2 m/s, and it is at dis-

tance d11.5 m from point O.

Particle P2has mass 3.1 kg and speed

v23.6 m/s, and it is at distance d2

2.8 m from point O. What are the

(a) magnitude and (b) direction of the

net angular momentum of the two particles about O?

••30 At the instant the displacement of a 2.00 kg object relative

to the origin is its veloc-

ity is and it is sub-

ject to a force Find (a) the

acceleration of the object, (b) the angular momentum of the object

about the origin, (c) the torque about the origin acting on the ob-

ject, and (d) the angle between the velocity of the object and the

force acting on the object.

••31 In Fig. 11-42, a 0.400 kg ball is

shot directly upward at initial speed 40.0

m/s. What is its angular momentum

about P, 2.00 m horizontally from the

launch point, when the ball is (a) at

maximum height and (b) halfway back

to the ground? What is the torque on the ball about Pdue to the

gravitational force when the ball is (c) at maximum height and

(d) halfway back to the ground?

Module 11-6 Newton’s Second Law in Angular Form

•32 A particle is acted on by two torques about the origin:



(4.00 N)k

ˆ.(8.00 N)j

ˆ

:(6.00 N)i

ˆ

(3.00 m/s)k

:(6.00 m/s)i

ˆ(3.00 m/s)j

ˆ

:(2.00 m)i

ˆ(4.00 m)j

ˆ(3.00 m)k

ˆ,

ILW

:(5.0i

ˆ5.0k

ˆ) m/s

:(2.0i

ˆ2.0k

ˆ) m

Figure 11-40

Problem 26.

Figure 11-41 Problem 29.

Od2

d1v2

Ball

Figure 11-42 Problem 31.

has a magnitude of 2.0 Nm and is directed in the positive direc-

tion of the xaxis, and has a magnitude of 4.0 N m and is

directed in the negative direction of the yaxis. In unit-vector

notation, find where is the angular momentum of the

particle about the origin.

•33 At time t0, a 3.0 kg particle with velocity

ILWWWWSSM



d

:/dt,

angle u330, acts on P. All three vectors lie in the xy plane.

About the origin, what are the (a) magnitude and (b) direction of

the angular momentum of Pand the (c) magnitude and (d) direc-

tion of the torque acting on P?

•27 At one instant, force acts on a 0.25 kg objectF

:4.0j

ˆ N

SSM

is at x3.0 m, y8.0 m. It is pulled

by a 7.0 N force in the negative xdirection.About the origin, what are

(a) the particle’s angular momentum,(b) the torque acting on the par-

ticle,and (c) the rate at which the angular momentum is changing?

•34 A particle is to move in an xy plane, clockwise around the

origin as seen from the positive side of the zaxis.In unit-vector nota-

tion, what torque acts on the particle if the magnitude of its angular

momentum about the origin is (a) 4.0 kgm2/s, (b) 4.0t2kg m2/s,

m/s)j

:(5.0 m/s)i

ˆ(6.0

323

PROBLEMS

Module 11-8 Conservation of Angular Momentum

•43 In Fig. 11-47, two skaters, each

of mass 50 kg, approach each other

along parallel paths separated by

3.0 m. They have opposite velocities

of 1.4 m/s each. One skater carries

one end of a long pole of negligible

mass, and the other skater grabs the

other end as she passes. The skaters

then rotate around the center of the pole. Assume that the friction

between skates and ice is negligible. What are (a) the radius of the

circle, (b) the angular speed of the skaters, and (c) the kinetic energy

of the two-skater system? Next, the skaters pull along the pole until

they are separated by 1.0 m. What then are (d) their angular speed

and (e) the kinetic energy of the system? (f) What provided the en-

ergy for the increased kinetic energy?

•44 A Texas cockroach of mass 0.17 kg runs counterclockwise

around the rim of a lazy Susan (a circular disk mounted on a vertical

axle) that has radius 15 cm, rotational inertia 5.0 103kg m2, and

frictionless bearings. The cockroach’s speed (relative to the ground)

is 2.0 m/s,and the lazy Susan turns clockwise with angular speed v0

2.8 rad/s. The cockroach finds a bread crumb on the rim and, of

course, stops. (a) What is the angular speed of the lazy Susan after the

cockroach stops? (b) Is mechanical energy conserved as it stops?

•45 A man stands on a platform that is rotating (with-

out friction) with an angular speed of 1.2 rev/s; his arms are

outstretched and he holds a brick in each hand. The rotational iner-

tia of the system consisting of the man, bricks, and platform about

the central vertical axis of the platform is 6.0 kgm2.If by moving the

bricks the man decreases the rotational inertia of the system to 2.0

kgm2, what are (a) the resulting angular speed of the platform and

(b) the ratio of the new kinetic energy of the system to the original

kinetic energy? (c) What source provided the added kinetic energy?

•46 The rotational inertia of a collapsing spinning star drops to

its initial value. What is the ratio of the new rotational kinetic en-

ergy to the initial rotational kinetic energy?

WWWSSM

•37 In Fig. 11-44, three particles

of mass m23 g are fastened to

three rods of length d12 cm and

negligible mass. The rigid assembly

rotates around point Oat the angu-

lar speed v0.85 rad/s. About O,

what are (a) the rotational inertia

of the assembly, (b) the magnitude

of the angular momentum of the middle particle, and (c) the mag-

nitude of the angular momentum of the asssembly?

•38 A sanding disk with rotational inertia 1.2 103kg m2is

attached to an electric drill whose motor delivers a torque of mag-

nitude 16 Nm about the central axis of the disk. About that axis

and with the torque applied for 33 ms, what is the magnitude of the

(a) angular momentum and (b) angular velocity of the disk?

•39 The angular momentum of a flywheel having a rota-

tional inertia of 0.140 kg m2about its central axis decreases from

3.00 to 0.800 kgm2/s in 1.50 s. (a) What is the magnitude of the av-

erage torque acting on the flywheel about its central axis during

this period? (b) Assuming a constant angular acceleration, through

what angle does the flywheel turn? (c) How much work is done on

the wheel? (d) What is the average power of the flywheel?

••40 A disk with a rotational inertia of rotates like

a merry-go-round while undergoing a time-dependent torque

given by .At

time 1.00 s, its angular momen-

tum is . What is its an-

gular momentum at s?

••41 Figure 11-45 shows a rigid

structure consisting of a circular

hoop of radius Rand mass m, and a

square made of four thin bars, each

of length Rand mass m. The rigid

structure rotates at a constant speed

about a vertical axis,with a period of

t3.00

5.00 kgm2/s

t

(5.00 2.00t) Nmt

7.00 kgm2



SSM



••35 At time t, the vector gives the

position of a 3.0 kg particle relative to the origin of an xy coordinate

system ( is in meters and tis in seconds). (a) Find an expression for

the torque acting on the particle relative to the origin. (b) Is the

magnitude of the particle’s angular momentum relative to the origin

increasing,decreasing,or unchanging?

Module 11-7 Angular Momentum of a Rigid Body

•36 Figure 11-43 shows three rotating, uniform disks that are cou-

pled by belts.One belt runs around the rims of disks Aand C.Another

belt runs around a central hub on disk Aand the rim of disk B.The

belts move smoothly without slippage on the rims and hub.Disk Ahas

radius R; its hub has radius 0.5000R; disk Bhas radius 0.2500R; and

disk Chas radius 2.000R. Disks Band Chave the same density (mass

per unit volume) and thickness.What is the ratio of the magnitude of

the angular momentum of disk Cto that of disk B?

:4.0t2i

ˆ(2.0t6.0t2)j

ˆrotation of 2.5 s. Assuming R0.50 m and m2.0 kg, calculate

(a) the structure’s rotational inertia about the axis of rotation and

(b) its angular momentum about that axis.

••42 Figure 11-46 gives the torque tthat acts on an initially stationary

disk that can rotate about its center like a merry-go-round. The scale

on the taxis is set by ts 4.0 Nm.What is the angular momentum of

the disk about the rotation axis at times (a) t7.0 s and (b) t20 s?

Belt

Rotation axis

R2R

Figure 11-43 Problem 36.

Figure 11-44 Problem 37.

Figure 11-45 Problem 41.

04 8 12 16 20

t (s)

(N • m)

Figure 11-46 Problem 42.

Figure 11-47 Problem 43.

324 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

•47 A track is mounted on a

large wheel that is free to turn with

negligible friction about a vertical

axis (Fig. 11-48). A toy train of mass

mis placed on the track and, with the

system initially at rest, the train’s

electrical power is turned on. The train reaches speed 0.15 m/s with

respect to the track. What is the wheel’s angular speed if its mass is

1.1mand its radius is 0.43 m? (Treat it as a hoop, and neglect the

mass of the spokes and hub.)

•48 A Texas cockroach walks from

the center of a circular disk (that ro-

tates like a merry-go-round without

external torques) out to the edge at

radius R. The angular speed of the

cockroach–disk system for the walk is

given in Fig. 11-49 (va= 5.0 rad/s and

vb= 6.0 rad/s). After reaching R,

what fraction of the rotational inertia

of the disk does the cockroach have?

•49 Two disks are mounted (like a merry-go-round) on low-

friction bearings on the same axle and can be brought together so

that they couple and rotate as one unit. The first disk, with rota-

tional inertia 3.30 kg m2about its central axis, is set spinning coun-

terclockwise at 450 rev/min. The second disk, with rotational inertia

6.60 kgm2about its central axis, is set spinning counterclockwise

at 900 rev/min.They then couple together. (a) What is their angular

speed after coupling? If instead the second disk is set spinning

clockwise at 900 rev/min, what are their (b) angular speed and

•50 The rotor of an electric motor has rotational inertia Im

2.0 103kg m2about its central axis. The motor is used to

change the orientation of the space probe in which it is mounted.

The motor axis is mounted along the central axis of the probe; the

probe has rotational inertia Ip12 kgm2about this axis.

Calculate the number of revolutions of the rotor required to turn

the probe through 30about its central axis.

•51 A wheel is rotating freely at angular speed

800 rev/min on a shaft whose rotational inertia is negligible. A sec-

ond wheel, initially at rest and with twice the rotational inertia of the

first, is suddenly coupled to the same shaft. (a) What is the angular

speed of the resultant combination of the shaft and two wheels?

(b) What fraction of the original rotational kinetic energy is lost?

••52 A cockroach of mass mlies on the rim of a uniform disk of

mass 4.00mthat can rotate freely about its center like a merry-go-

round. Initially the cockroach and disk rotate together with an angu-

lar velocity of 0.260 rad/s. Then the cockroach walks halfway to the

center of the disk. (a) What then is the angular velocity of the cock-

roach–disk system? (b) What is the ratio K/K0of the new kinetic en-

ergy of the system to its initial kinetic energy? (c) What accounts for

the change in the kinetic energy?

••53 In Fig. 11-50 (an overhead

view), a uniform thin rod of length

0.500 m and mass 4.00 kg can rotate

in a horizontal plane about a verti-

cal axis through its center. The rod is

at rest when a 3.00 g bullet traveling

in the rotation plane is fired into one

end of the rod. In the view from

ILWSSM

SSM above, the bullet’s path makes angle u60.0with the rod (Fig. 11-

50). If the bullet lodges in the rod and the angular velocity of the rod

is 10 rad/s immediately after the collision, what is the bullet’s speed

just before impact?

••54 Figure 11-51 shows an

overhead view of a ring that can

rotate about its center like a merry-

go-round. Its outer radius R2is

0.800 m, its inner radius R1is R2/2.00,

its mass Mis 8.00 kg, and the mass of

the crossbars at its center is neg-

ligible. It initially rotates at an angu-

lar speed of 8.00 rad/s with a cat of

mass mM/4.00 on its outer edge, at

radius R2. By how much does the cat increase the kinetic energy of

the cat–ring system if the cat crawls to the inner edge, at radius R1?

••55 A horizontal vinyl record of mass 0.10 kg and radius 0.10 m

rotates freely about a vertical axis through its center with an angu-

lar speed of 4.7 rad/s and a rotational inertia of 5.0 104kgm2.

Putty of mass 0.020 kg drops vertically onto the record from above

and sticks to the edge of the record. What is the angular speed of

the record immediately afterwards?

••56 In a long jump, an athlete leaves the ground with an

initial angular momentum that tends to rotate her body forward,

threatening to ruin her landing.To counter this tendency, she rotates

her outstretched arms to “take up” the angular momentum (Fig. 11-

18). In 0.700 s, one arm sweeps through 0.500 rev and the other arm

sweeps through 1.000 rev.Treat each arm as a thin rod of mass 4.0 kg

and length 0.60 m, rotating around one end. In the athlete’s reference

frame, what is the magnitude of the total angular momentum of the

arms around the common rotation axis through the shoulders?

••57 A uniform disk of mass 10mand radius 3.0rcan rotate

freely about its fixed center like a merry-go-round.A smaller uni-

form disk of mass mand radius rlies on top of the larger disk,

concentric with it. Initially the two disks rotate together with an an-

gular velocity of 20 rad/s.Then a slight disturbance causes the smaller

disk to slide outward across the larger disk, until the outer edge of the

smaller disk catches on the outer edge of the larger disk. Afterward,

the two disks again rotate together (without further sliding).(a) What

then is their angular velocity about the center of the larger disk? (b)

What is the ratio K/K0of the new kinetic energy of the two-disk sys-

tem to the system’s initial kinetic energy?

••58 A horizontal platform in the shape of a circular disk rotates on

a frictionless bearing about a vertical axle through the center of the

disk. The platform has a mass of 150 kg, a radius of 2.0 m, and a rota-

tional inertia of 300 kgm2about the axis of rotation.A 60 kg student

walks slowly from the rim of the platform toward the center.If the an-

gular speed of the system is 1.5 rad/s when the student starts at the

rim,what is the angular speed when she is 0.50 m from the center?

••59 Figure 11-52 is an overhead

view of a thin uniform rod of length

0.800 m and mass Mrotating horizon-

tally at angular speed 20.0 rad/s about

an axis through its center. A particle

of mass M/3.00 initially attached to

one end is ejected from the rod and travels along a path that is per-

pendicular to the rod at the instant of ejection. If the particle’s speed

vpis 6.00 m/s greater than the speed of the rod end just after ejec-

tion, what is the value of vp?

Figure 11-48 Problem 47.

(rad/s)

Radial distance R

Figure 11-49 Problem 48.

Axis

Figure 11-51 Problem 54.

Figure 11-50 Problem 53.

Rotation

axis

Figure 11-52 Problem 59.

325

PROBLEMS

••60 In Fig. 11-53, a 1.0 g bullet is fired

into a 0.50 kg block attached to the end

of a 0.60 m nonuniform rod of mass

0.50 kg. The block–rod–bullet system

then rotates in the plane of the figure,

about a fixed axis at A. The rotational

inertia of the rod alone about that axis

at Ais 0.060 kgm2.Treat the block as a

particle. (a) What then is the rotational

inertia of the block–rod–bullet system

about point A? (b) If the angular speed

of the system about Ajust after impact

is 4.5 rad/s, what is the bullet’s speed

just before impact?

••61 The uniform rod (length 0.60 m,

mass 1.0 kg) in Fig. 11-54 rotates in the

plane of the figure about an axis through

one end, with a rotational inertia of 0.12

kgm2. As the rod swings through its

lowest position, it collides with a 0.20 kg

putty wad that sticks to the end of the

rod.If the rod’s angular speed just before

collision is 2.4 rad/s, what is the angular

speed of the rod–putty system immedi-

ately after collision?

•••62 During a jump to his partner, an aerialist is to

tangent to the outer edge of the merry-go-round, as shown. What is

the angular speed of the merry-go-round just after the ball is caught?

•••64 A ballerina begins a tour jeté (Fig. 11-19a) with angu-

lar speed and a rotational inertia consisting of two parts:

for her leg extended outward at angle

to her body and for the rest of her body (pri-

marily her trunk). Near her maximum height she holds both legs at

angle to her body and has angular speed (Fig. 11-19b).

Assuming that has not changed, what is the ratio ?

•••65 Two 2.00 kg balls are attached to the ends of a

thin rod of length 50.0 cm and negli-

gible mass.The rod is free to rotate in

a vertical plane without friction

about a horizontal axis through its

center.With the rod initially horizon-

tal (Fig. 11-57), a 50.0 g wad of wet

putty drops onto one of the balls, hit-

ting it with a speed of 3.00 m/s and then sticking to it. (a) What is the

angular speed of the system just after the putty wad hits? (b) What is

the ratio of the kinetic energy of the system after the collision to that

of the putty wad just before? (c) Through what angle will the system

rotate before it momentarily stops?

WWWSSM

vf/vi

Itrunk

u30.0

Itrunk 0.660 kgm2

u90.0Ileg 1.44 kgm2

Bullet

Block

Rod

Figure 11-53 Problem 60.

Figure 11-54 Problem 61.

Rod

Rotation axis

Release

Catch

Parabolic

path of

aerialist

Tuck

Figure 11-55 Problem 62.

Figure 11-57 Problem 65.

Putty wad

Rotation

axis

Figure 11-58 Problem 66.

Figure 11-59 Problem 67.

•••63 In Fig. 11-56, a 30 kg child

stands on the edge of a stationary

merry-go-round of radius 2.0 m.

The rotational inertia of the merry-

go-round about its rotation axis is

150 kgm2. The child catches a ball

of mass 1.0 kg thrown by a friend.

Just before the ball is caught, it has a

horizontal velocity of magnitude

12 m/s, at angle f37with a line

Figure 11-56 Problem 63.

Child

Ball

make a quadruple somersault lasting a time t 1.87 s. For the first

and last quarter-revolution, he is in the extended orientation

shown in Fig. 11-55, with rotational inertia I1 19.9 kgm2around

his center of mass (the dot). During the rest of the flight he is in a

tight tuck, with rotational inertia I2 3.93 kgm2.What must be his

angular speed v2around his center of mass during the tuck?

•••66 In Fig. 11-58, a small 50 g

block slides down a frictionless sur-

face through height h20 cm and

then sticks to a uniform rod of mass

100 g and length 40 cm.The rod pivots

about point Othrough angle u

before momentarily stopping. Find u.

•••67 Figure 11-59 is an over-

head view of a thin uniform rod of

length 0.600 m and mass Mrotating

horizontally at 80.0 rad/s counter-

clockwise about an axis through its center. A particle of mass

M/3.00 and traveling horizontally at speed 40.0 m/s hits the rod

and sticks. The particle’s path is perpendicular to the rod at the

instant of the hit, at a distance dfrom the rod’s center. (a) At

what value of dare rod and particle stationary after the hit?

(b) In which direction do rod and particle rotate if dis greater

than this value?

Module 11-9 Precession of a Gyroscope

••68 A top spins at 30 rev/s about an axis that makes an angle of

30with the vertical.The mass of the top is 0.50 kg, its rotational in-

ertia about its central axis is 5.0 104kgm2, and its center of

mass is 4.0 cm from the pivot point. If the spin is clockwise from an

overhead view, what are the (a) precession rate and (b) direction of

the precession as viewed from overhead?

••69 A certain gyroscope consists of a uniform disk with a 50 cm

radius mounted at the center of an axle that is 11 cm long and of

negligible mass.The axle is horizontal and supported at one end. If

the spin rate is 1000 rev/min, what is the precession rate?

Rotation axis Particle

326 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM

Additional Problems

70 A uniform solid ball rolls smoothly along a floor, then up a

ramp inclined at 15.0. It momentarily stops when it has rolled

1.50 m along the ramp.What was its initial speed?

71 In Fig. 11-60, a constant

horizontal force of magnitude 12

N is applied to a uniform solid cylin-

der by fishing line wrapped around

the cylinder.The mass of the cylinder

is 10 kg, its radius is 0.10 m, and the

cylinder rolls smoothly on the hori-

zontal surface. (a) What is the mag-

nitude of the acceleration of the center of mass of the cylinder? (b)

What is the magnitude of the angular acceleration of the cylinder

about the center of mass? (c) In unit-vector notation, what is the

frictional force acting on the cylinder?

72 A thin-walled pipe rolls along the floor.What is the ratio of its

translational kinetic energy to its rotational kinetic energy about

the central axis parallel to its length?

73 A 3.0 kg toy car moves along an xaxis with a velocity

given by with tin seconds. For t0, what are

(a) the angular momentum of the car and (b) the torque on

the car, both calculated about the origin? What are (c) and (d)

about the point (2.0 m, 5.0 m, 0)? What are (e) and (f) about

the point (2.0 m, 5.0 m, 0)?

74 A wheel rotates clockwise about its central axis with an angu-

lar momentum of 600 kgm2/s. At time t0, a torque of magni-

tude 50 Nm is applied to the wheel to reverse the rotation. At

what time tis the angular speed zero?

75 In a playground, there is a small merry-go-round of

radius 1.20 m and mass 180 kg. Its radius of gyration (see Problem

79 of Chapter 10) is 91.0 cm.A child of mass 44.0 kg runs at a speed

of 3.00 m/s along a path that is tangent to the rim of the initially

stationary merry-go-round and then jumps on. Neglect friction be-

tween the bearings and the shaft of the merry-go-round. Calculate

(a) the rotational inertia of the merry-go-round about its axis of

rotation, (b) the magnitude of the angular momentum of the run-

ning child about the axis of rotation of the merry-go-round, and

child has jumped onto the merry-go-round.

76 A uniform block of granite in the shape of a book has face di-

mensions of 20 cm and 15 cm and a thickness of 1.2 cm. The density

(mass per unit volume) of granite is 2.64 g/cm3. The block rotates

around an axis that is perpendicular to its face and halfway between

its center and a corner. Its angular momentum about that axis is

0.104 kgm2/s. What is its rotational kinetic energy about that axis?

77 Two particles, each of mass 2.90 104kg and speed

5.46 m/s, travel in opposite directions along parallel lines separated

by 4.20 cm. (a) What is the magnitude Lof the angular momentum

of the two-particle system around a point midway between the two

lines? (b) Is the value different for a different location of the

point? If the direction of either particle is reversed, what are the

answers for (c) part (a) and (d) part (b)?

78 A wheel of radius 0.250 m, moving initially at 43.0 m/s, rolls to

a stop in 225 m. Calculate the magnitudes of its (a) linear accelera-

tion and (b) angular acceleration. (c) Its rotational inertia is 0.155

kgm2about its central axis. Find the magnitude of the torque

about the central axis due to friction on the wheel.

SSM

:v

:2.0t3i

ˆ m/s,

SSM

app

SSM

79 Wheels Aand Bin Fig. 11-61 are

connected by a belt that does not slip.

The radius of Bis 3.00 times the radius

of A. What would be the ratio of the

rotational inertias IA/IBif the two

wheels had (a) the same angular mo-

mentum about their central axes and

(b) the same rotational kinetic energy?

80 A 2.50 kg particle that is moving horizontally over a floor

with velocity (3.00 m/s)ˆ

j undergoes a completely inelastic colli-

sion with a 4.00 kg particle that is moving horizontally over the

floor with velocity (4.50 m/s)i

ˆ. The collision occurs at xy coordi-

nates (0.500 m, 0.100 m). After the collision and in unit-vector

notation, what is the angular momentum of the stuck-together par-

ticles with respect to the origin?

81 A uniform wheel

of mass 10.0 kg and radius

0.400 m is mounted rigidly

on a massless axle through

its center (Fig. 11-62). The

radius of the axle is 0.200

m, and the rotational inertia

of the wheel–axle combi-

nation about its central axis

is 0.600 kgm2.The wheel is

initially at rest at the top of a surface that is inclined at angle u

30.0with the horizontal; the axle rests on the surface while the

wheel extends into a groove in the surface without touching the

surface. Once released, the axle rolls down along the surface

smoothly and without slipping.When the wheel–axle combination

has moved down the surface by 2.00 m, what are (a) its rotational

kinetic energy and (b) its translational kinetic energy?

82 A uniform rod rotates in a horizontal plane about a vertical axis

through one end.The rod is 6.00 m long,weighs 10.0 N, and rotates at

240 rev/min. Calculate (a) its rotational inertia about the axis of rota-

tion and (b) the magnitude of its angular momentum about that axis.

83 A solid sphere of weight 36.0 N rolls up an incline at an angle

of 30.0. At the bottom of the incline the center of mass of the

sphere has a translational speed of 4.90 m/s. (a) What is the kinetic

energy of the sphere at the bottom of the incline? (b) How far does

the sphere travel up along the incline? (c) Does the answer to

(b) depend on the sphere’s mass?

84 Suppose that the yo-yo in Problem 17, instead of rolling

from rest, is thrown so that its initial speed down the string is

1.3 m/s. (a) How long does the yo-yo take to reach the end of

the string? As it reaches the end of the string, what are its (b) total

kinetic energy, (c) linear speed, (d) translational kinetic energy,

(e) angular speed, and (f) rotational kinetic energy?

85 A girl of mass Mstands on the rim of a frictionless merry-

go-round of radius Rand rotational inertia Ithat is not moving. She

throws a rock of mass mhorizontally in a direction that is tangent to

the outer edge of the merry-go-round.The speed of the rock, relative

to the ground, is v. Afterward, what are (a) the angular speed of the

merry-go-round and (b) the linear speed of the girl?

86 A body of radius Rand mass mis rolling smoothly with speed

von a horizontal surface. It then rolls up a hill to a maximum

height h. (a) If h3v2/4g, what is the body’s rotational inertia

about the rotational axis through its center of mass? (b) What

might the body be?

SSM

Figure 11-60 Problem 71.

Fishing line

Fapp

Figure 11-62 Problem 81.

Groove

Axle

Wheel

Figure 11-61 Problem 79.